Math Genius: Proving convergence of improper integral, g(t)

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I want to prove convergence of the improper integral,

$$int_1^infty g(t) dt$$
Where $g(t)$ is given by: $g(t)=(frac{1-e^{-2t^2}}{t^2})$ , $tneq0$

I have thought about using different tests, but I keep getting stuck at the fact that as t approaches zero from the right hand side, the function value increases towards infinity, so I can’t figure out a limit from the right hand side.

I hope you can help.

Best regards,
Christoffer

$0<frac{1−e^{−t^2}}{t^2} < frac{1}{t^2}$ which is integrable on $[1,∞)$, so by comparison, your integral converges.

If you look at the integrand, it can be rewritten as
$$
f(t) = 2 int_{0}^{1} e^{-2t^2x} dx
$$

So the whole expression (with $varepsilon to infty$)
$$
L = 2 int_{1}^{varepsilon}int_{0}^{1}e^{-2xt^2}dxdt = 2int_{0}^{1}int_{1}^{varepsilon} e^{-2xt^2}dt dx
$$

Use the $texttt{erf}$ function for the second integral:
$$
L = sqrt{2} int_{0}^{1}frac{1}{sqrt{x}}[texttt{erf}(varepsilon) – texttt{erf}(1)]
$$

Now, compute the simple integral, and then take the limit as $varepsilon to infty$. Can you handle from here?

We use the special function $$text {erf}(x)=sqrt{frac{2}{pi}}int_0^xe^{-t^2}dt$$ and the fact that $lim_{x to infty}text{erf}(x)=1.$ First we integrate from 1 to $b$ and then we take the limit as $b to infty$. Integration by parts gives $$int_1^bfrac{1-e^{-2t^2}}{t^2}dt=int_1^b(-frac{1}{t})'(1-e^{-2t^2})$$
$$=frac{-(1-e^{-2t^2})}{t}|_1^b+4int_1^be^{-2t^2}dt$$
$$=frac{-(1-e^{-2b^2})}{b}+1-e^{-2}+frac{4}{sqrt 2}int_{sqrt 2}^{sqrt 2 b}e^{-u^2}du$$ after making the substitution $u=sqrt 2 t$. Thus the inegral is
$$frac{-(1-e^{-2b^2})}{b}+1-e^{-2}+frac{4}{sqrt 2}(int_0^{sqrt 2 b}e^{-u^2}du-int_0^{sqrt 2}e^{-u^2}du)$$ $$=frac{-(1-e^{-2b^2})}{b}+1-e^{-2}+2sqrt{pi}(text{erf}(sqrt 2 b)-text{erf}(sqrt 2))$$ Taking the limit as $b to infty$, the given integral converges to $$1-e^{-2}+2sqrtpi(1-text{erf}(sqrt 2)).$$

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Math Genius: What is a method for solving principal value integral of $frac{1}{pi}int_{-B}^{B} frac{x sqrt{B^2-x^2}}{x-y}mathrm{d} x$?

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Question:

I am trying to solve a principal value integral involving a square root. Using Mathematica I can get an answer but I would like to know a general approach to obtain them by hand. To be clear I am interested in a clear explanation of the method not just the solution.

The principal value integral is of the form:
$$ I_{B}left(yright)=frac{1}{pi} int_0^B Jleft(xright)left(frac{mathcal{P}}{x+y}+frac{mathcal{P}}{x-y}right)mathrm{d}x,$$
where $mathcal{P}$ denotes a principal value integral and $0le yle B$ (we only consider real numbers). I would be interested in a method of solution for $Jleft(xright)$ given by:
$$Jleft(xright)=xsqrt{B^2-x^2}.$$

Using Mathematica I have ascertained that $Ileft(yright)=frac{B^2}{2}-y^2$ and since $Jleft(xright)$ is odd we have:
$$ I_{B}left(yright)=frac{1}{pi} int_{-B}^B Jleft(xright)frac{mathcal{P}}{x-y}mathrm{d}x,$$
so I wonder if it could be solved with something akin to contour integration but I am at a loss as to how to proceed further.

Context:

The integral is required to map spectral densities of Hamiltonians used in Open Quantum Systems.

Bonus:

I will accept any answer that provides a step-by-step (analytic) method of solution for $I$ but I would also be interested in methods of solution for a couple of other integrals. I can post these as a separate question if people prefer.

Firstly:
$$ L_{left(C,Dright)}left(yright)=frac{1}{pi} int_{C}^{D} J_2left(xright)frac{mathcal{P}}{x-y}mathrm{d}x,$$
with $0le C<y<D$, all positive real numbers and:
$$J_2left(xright)=sqrt{left(D-xright)left(x-Cright)}.$$
Unfortunately, I have been unable to compute $L$ with Mathematica, but I have used alternative methods to ascertain $L_{left(C,Dright)}left(yright)=frac{C+D}{2}-y$.

And also:
$$ K_{left(A,Bright)}left(yright)=frac{1}{pi} int_{A}^{B} J_1left(xright)left(frac{mathcal{P}}{x+y}+frac{mathcal{P}}{x-y}right)mathrm{d}x,$$
with $0le A<y<B$ all positive real numbers and:
$$J_1left(xright)=sqrt{left(B^2-x^2right)left(x^2-A^2right)}.$$
Since:
$$ frac{1}{pi} int_{A}^{B} J_1left(xright)left(frac{1}{x+y}+frac{1}{x-y}right)mathrm{d}x=frac{1}{pi} int_{A}^{B} J_1left(xright)frac{2x}{x^2-y^2}mathrm{d}x=frac{1}{pi} int_{A^2}^{B^2} J_1left(sqrt{w}right)frac{1}{w-y^2}mathrm{d}x=Lleft(y^2right)=frac{A^2+B^2}{2}-y^2,$$
where we have used the substitution $w=x^2$ and defined $C=A^2$ and $D=B^2$.

It is probably also worth noting that:
$$ L_{left(C,Dright)}left(yright)=frac{1}{pi} int_{0}^{D-C} sqrt{w}sqrt{D-C-w}frac{mathcal{P}}{w-(y-C)}mathrm{d}w,$$
which can be found using the substitution $w=x-C$.

The evaluation of the Cauchy principal value integral via contour integration is relatively straightforward. To begin, consider the contour integral

$$oint_C dz , frac{z sqrt{z^2-B^2}}{z-y} $$

where $C$ is the following contour:

enter image description here

There are semicircular detours of radius $epsilon$ around the branch points at $z=pm B$ and the pole at $z=y$. Also, the large circle has a radius $R$. The pieces of the contour integral as labeled in the figure are as follows. (Yes, there are a lot of pieces, but as you will see, most will vanish or cancel.)

$$int_{AB} dz frac{z sqrt{z^2-B^2}}{z-y} = int_{-R}^{-B-epsilon} dx frac{x sqrt{x^2-B^2}}{x-y}$$

$$int_{BC} dz frac{z sqrt{z^2-B^2}}{z-y} = i epsilon int_{pi}^0 dphi , e^{i phi} , frac{(-B+epsilon e^{i phi})sqrt{(-B+epsilon e^{i phi})^2-B^2}}{-B+epsilon e^{i phi}-y} $$

$$int_{CD} dz frac{z sqrt{z^2-B^2}}{z-y} = int_{-B+epsilon}^{y-epsilon} dx frac{x sqrt{x^2-B^2}}{x-y} = int_{-B+epsilon}^{y-epsilon} dx frac{x i sqrt{B^2-x^2}}{x-y}$$

$$int_{DE} dz frac{z sqrt{z^2-B^2}}{z-y} = i epsilon int_{pi}^0 dphi , e^{i phi} , frac{(y+epsilon e^{i phi})sqrt{(y+epsilon e^{i phi})^2-B^2}}{epsilon e^{i phi}} \ = i epsilon int_{pi}^0 dphi , e^{i phi} , frac{(y+epsilon e^{i phi})i sqrt{B^2-(y+epsilon e^{i phi})^2}}{epsilon e^{i phi}}$$

$$int_{EF} dz frac{z sqrt{z^2-B^2}}{z-y} = int_{y+epsilon}^{B-epsilon} dx frac{x sqrt{x^2-B^2}}{x-y} = int_{y+epsilon}^{B-epsilon} dx frac{x i sqrt{B^2-x^2}}{x-y}$$

$$int_{FG} dz frac{z sqrt{z^2-B^2}}{z-y} = i epsilon int_{pi}^{-pi} dphi , e^{i phi} , frac{(B+epsilon e^{i phi}) sqrt{(B+epsilon e^{i phi})^2-B^2}}{B+epsilon e^{i phi}-y} $$

$$int_{GH} dz frac{z sqrt{z^2-B^2}}{z-y} = int_{B-epsilon}^{y+epsilon} dx frac{x sqrt{x^2-B^2}}{x-y} = int_{B-epsilon}^{y+epsilon} dx frac{x (-i) sqrt{B^2-x^2}}{x-y}$$

$$int_{HI} dz frac{z sqrt{z^2-B^2}}{z-y} = i epsilon int_{0}^{-pi} dphi , e^{i phi} , frac{(y+epsilon e^{i phi})sqrt{(y+epsilon e^{i phi})^2-B^2}}{epsilon e^{i phi}} \ = i epsilon int_{0}^{-pi} dphi , e^{i phi} , frac{(y+epsilon e^{i phi})(-i) sqrt{B^2-(y+epsilon e^{i phi})^2}}{epsilon e^{i phi}} $$

$$int_{IJ} dz frac{z sqrt{z^2-B^2}}{z-y} = int_{y-epsilon}^{-B+epsilon} dx frac{x sqrt{x^2-B^2}}{x-y} = int_{y-epsilon}^{-B+epsilon} dx frac{x (-i) sqrt{B^2-x^2}}{x-y}$$

$$int_{JK} dz frac{z sqrt{z^2-B^2}}{z-y} = i epsilon int_{0}^{-pi} dphi , e^{i phi} , frac{(-B+epsilon e^{i phi})sqrt{(-B+epsilon e^{i phi})^2-B^2}}{-B+epsilon e^{i phi}-y} $$

$$int_{KL} dz frac{z sqrt{z^2-B^2}}{z-y} = int_{-B-epsilon}^{-R} dx frac{x sqrt{x^2-B^2}}{x-y}$$

$$int_{LA} dz frac{z sqrt{z^2-B^2}}{z-y} = i R int_{-pi}^{pi} dtheta , e^{i theta} frac{R e^{i theta} sqrt{R^2 e^{i 2 theta}-B^2}}{R e^{i theta}-y} $$

Note that, on the branch above the real axis, $-1=e^{i pi}$ and on the branch below the real axis, $-1=e^{-i pi}$. Thus, the sign of $i$ in front of the square root when $|x| lt B$ is positive above the real axis and negative below the real axis.

First, note that the integrals over $AB$ and $KL$ cancel because the square root does not introduce any phase change there.

Second, as $epsilon to 0$, the integrals over $BC$, $FG$, and $JK$ vanish.

Third, as $epsilon to 0$, the integrals over $DE$ and $HI$ cancel. In this case, the phase difference over the branch cut due to the square root turns what is normally a constructive interference (i.e., the contributions usually add) into a destructive interference (i.e., they cancel.)

Fourth, as $epsilon to 0$, the integrals over $CD$ and $EF$ combine to form

$$i PV int_{-B}^B dx frac{x sqrt{B^2-x^2}}{x-y}$$

and the integrals over $GH$ and $IJ$ combine to form

$$-i PV int_{B}^{-B} dx frac{x sqrt{B^2-x^2}}{x-y}$$

so together, the contribution to the contour integral over these four intervals is

$$i 2 PV int_{-B}^B dx frac{x sqrt{B^2-x^2}}{x-y}$$

Fifth, the final contribution to the contour integral is the integral over $LA$ in the limit as $R to infty$. That limit is evaluated as follows:

$$begin{align} int_{LA} dz frac{z sqrt{z^2-B^2}}{z-y} &= i R int_{-pi}^{pi} dtheta , e^{i theta} frac{R e^{i theta} sqrt{R^2 e^{i 2 theta}-B^2}}{R e^{i theta}-y} \ &= i R^2 int_{-pi}^{pi} dtheta , e^{i 2 theta} left (1-frac{B^2}{R^2 e^{i 2 theta}} right )^{1/2} left (1-frac{y}{R e^{i theta}} right )^{-1} \ &= i R^2 int_{-pi}^{pi} dtheta , e^{i 2 theta} left [1+y frac{1}{R e^{i theta}} -left (frac{B^2}{2} – y^2 right ) frac1{R^2 e^{i 2 theta}} + O left ( frac1{R^3} right ) right ]end{align}$$

After integration, the first two contributions inside the brackets vanish. Further, as $R to infty$, all terms $O(1/R^3)$ will also vanish. This simply leaves the $1/R^2$ term in the integrand, and we may finally write an expression for the contour integral:

$$oint_C dz , frac{z sqrt{z^2-B^2}}{z-y} = i 2 PV int_{-B}^B dx frac{x sqrt{B^2-x^2}}{x-y} – i 2 pi left (frac{B^2}{2} – y^2 right )$$

By Cauchy’s theorem, the contour integral is equal to zero. Therefore, when $y in (-B,B)$

$$frac1{pi} PV int_{-B}^B dx frac{x sqrt{B^2-x^2}}{x-y} = frac{B^2}{2} – y^2$$

as asserted by the OP.

ADDENDUM

The case $|y| gt B$ is a straightforward application of the residue theorem and the result is

$$frac1{pi} int_{-B}^B dx frac{x sqrt{B^2-x^2}}{x-y} = sqrt{y^2-B^2} left (y – sqrt{y^2-B^2} right ) – frac{B^2}{2}$$

Preliminary definitions and equivalences:

Firstly we define (as per the question):
begin{equation}
begin{split}
I_{B}left(yright)=&frac{1}{pi} int_{0}^{B} xsqrt{B^2-x^2}left(frac{mathcal{P}_{x+y}}{x+y}+frac{mathcal{P}_{x-y}}{x-y}right)mathrm{d}x,\
K_{left(A,Bright)}left(yright)=&frac{1}{pi} int_{A}^{B} sqrt{left(B^2-x^2right)left(x^2-A^2right)}left(frac{mathcal{P}_{x+y}}{x+y}+frac{mathcal{P}_{x-y}}{x-y}right)mathrm{d}x,\
L_{left(C,Dright)}left(yright)=&frac{1}{pi} int_{C}^{D} sqrt{left(D-xright)left(x-Cright)}frac{mathcal{P}_{x-y}}{x-y}mathrm{d}x,\
M_{E}left(yright)=&frac{1}{pi} int_{0}^{E} sqrt{x}sqrt{E-x}frac{mathcal{P}_{x-y}}{x-y}mathrm{d}x,\
Nleft(yright)=&frac{1}{pi} int_{0}^{1} sqrt{x}sqrt{1-x}frac{mathcal{P}_{x-y}}{x-y}mathrm{d}x.
end{split}
end{equation}

where $frac{mathcal{P}_{x}}{x}$ is a principal value distribution centred at $x=0$:
begin{equation}
begin{split}
frac{mathcal{P}_{x}}{x}=lim_{varepsilonrightarrow 0}left[frac{x}{x^2+varepsilon^2}right]underset{mathcal{D}}{=}begin{cases}
frac{1}{x}, &text{ for } xne 0\
0, &text{ for } x= 0\
end{cases}
end{split}
end{equation}

where the limit should be taken after integration, we will also define a distributional equality: $=_{mathcal{D}}$. If two distributions are distributionally equal this means they will have the same value when integrated over any domain (assuming there are no poles in the integrand other than those that appear explicitly in the distributional equality). If two distributions are distributionally equal this means they differ by a finite amount (in fact they can differ by an infinite amount as long as the infinity diverges slower than $1/x$ does, as $xrightarrow 0$) at a finite (or countably infinite) number of points within their domain. Distributional equality holds in such a case because (unless a distribution is infinitely valued at such a point) the contribution of a single point in an integration domain to the full integral is infinitesimal (zero). A commonly used distributionally equality is:
begin{equation}
begin{split}
xfrac{mathcal{P}_{x}}{x}underset{mathcal{D}}{=}1,
end{split}
end{equation}

since the two only differ at a single point $x=0$, where the former is zero and the later one (so they differ by a finite amount (1) at a finite number (1) of points).

In all the integrals above we assume that $y$ is within the integration domain (though it can be at a boundary), e.g. for $I_{B}left(yright)$ we have $0le yle B$ and for $K_{left(A,Bright)}left(yright)$ we have $Ale yle B$. Now we have $I_{B}left(yright)=K_{left(0,Bright)}left(yright)$ by definition;
begin{equation}
begin{split}
K_{left(A,Bright)}left(yright)=&frac{1}{pi} int_{A}^{B} sqrt{left(B^2-x^2right)left(x^2-A^2right)}left(frac{mathcal{P}_{x+y}}{x+y}+frac{mathcal{P}_{x-y}}{x-y}right)mathrm{d}x\
=&frac{1}{pi} int_{A}^{B} 2xsqrt{left(B^2-x^2right)left(x^2-A^2right)}frac{mathcal{P}_{x^2-y^2}}{x^2-y^2}mathrm{d}x\
=&frac{1}{pi} int_{A^2}^{B^2} sqrt{left(B^2-wright)left(w-A^2right)}frac{mathcal{P}_{w-y^2}}{w-y^2}mathrm{d}w=L_{left(A^2,B^2right)}left(y^2right),
end{split}
end{equation}

using the distributional equality [1]:
begin{equation}
begin{split}
frac{mathcal{P}_{x+y}}{x+y}+frac{mathcal{P}_{x-y}}{x-y}underset{mathcal{D}}{=}2xfrac{mathcal{P}_{x^2-y^2}}{x^2-y^2},
end{split}
end{equation}

and the substitution $w=x^2$ (so $2xmathrm{d} x=mathrm{d} w$); we also have:
begin{equation}
begin{split}
L_{left(C,Dright)}left(yright)=&frac{1}{pi} int_{C}^{D} sqrt{left(D-xright)left(x-Cright)}frac{mathcal{P}_{x-y}}{x-y}mathrm{d}x\
=&frac{1}{pi} int_{0}^{D-C} sqrt{left(D-C-wright)w}frac{mathcal{P}_{w-left(y-Cright)}}{w-left(y-Cright)}mathrm{d}w=M_{D-C}left(y-Cright),\
end{split}
end{equation}

where we use the substitution $w=x-C$; and lastly [2]:
begin{equation}
begin{split}
M_{E}left(yright)=&frac{1}{pi} int_{0}^{E} sqrt{x}sqrt{E-x}frac{mathcal{P}_{x-y}}{x-y}mathrm{d}x=frac{1}{pi} int_{0}^{1} sqrt{Ew}sqrt{E-Ew}frac{mathcal{P}_{Ew-y}}{Ew-y}Emathrm{d}w\
=&Efrac{1}{pi} int_{0}^{1} sqrt{w}sqrt{1-w}frac{mathcal{P}_{w-frac{y}{E}}}{w-frac{y}{E}}mathrm{d}w=E Nleft(frac{y}{E}right),\
end{split}
end{equation}

where we have used the substitution $w=frac{x}{E}$.

So we have:
begin{equation}
begin{split}
M_{E}left(yright)=&E Nleft(frac{y}{E}right),\
L_{left(C,Dright)}left(yright)=&M_{D-C}left(y-Cright)=left(D-Cright) Nleft(frac{y-C}{D-C}right)\
K_{left(A,Bright)}left(yright)=&L_{left(A^2,B^2right)}left(y^2right)=M_{B^2-A^2}left(y^2-A^2right)=left(B^2-A^2right) Nleft(frac{y^2-A^2}{B^2-A^2}right),\
I_{B}left(yright)=&K_{left(0,Bright)}left(yright)=B^2 Nleft(frac{y^2}{B^2}right),
end{split}
end{equation}

and only need to determine a method to integrate $Nleft(yright)$ to answer all the original questions.

Integrations:

If we consider $Nleft(yright)$ for $0<y<1$ we find:
begin{equation}
begin{split}
Nleft(yright)=&frac{1}{pi} int_{0}^{1} sqrt{x}sqrt{1-x}frac{mathcal{P}_{x-y}}{x-y}mathrm{d}x=frac{2}{pi} int_{0}^{1} w^2sqrt{1-w^2}frac{mathcal{P}_{w^2-y}}{w^2-y}mathrm{d}w\
=&frac{2}{pi} int_{0}^{1} left(w^2-y+yright)sqrt{1-w^2}frac{mathcal{P}_{w^2-y}}{w^2-y}mathrm{d}w\
=&frac{2}{pi} int_{0}^{1} sqrt{1-w^2}mathrm{d}w+frac{2y}{pi} int_{0}^{1} sqrt{1-w^2}frac{mathcal{P}_{w^2-y}}{w^2-y}mathrm{d}w\
=&frac{1}{pi} int_{0}^{frac{pi}{2}} left(1+cosleft(2thetaright)right)mathrm{d}theta+frac{2y}{pi} int_{0}^{frac{pi}{2}} left(1-sin^2theta+y-yright)frac{mathcal{P}_{sin^2theta-y}}{sin^2theta-y}mathrm{d}theta\
=&frac{1}{pi} left[theta+frac{1}{2}sinleft(2thetaright)right]_{0}^{frac{pi}{2}}-frac{2y}{pi} int_{0}^{frac{pi}{2}} left(sin^2theta-yright)frac{mathcal{P}_{sin^2theta-y}}{sin^2theta-y}mathrm{d}theta\
&+frac{2yleft(1-yright)}{pi} int_{0}^{frac{pi}{2}} frac{mathcal{P}_{sin^2theta-y}}{sin^2theta-y}mathrm{d}theta\
=&frac{1}{2}-y,
end{split}
end{equation}

where we have used the substitution $w=sqrt{x}$ (so $mathrm{d}x=2wmathrm{d}w$) and then $w=
sintheta$
(so $mathrm{d}w=costhetamathrm{d}theta$), we have also used the distributional equality:
begin{equation}
begin{split}
left(x-yright)frac{mathcal{P}_{x-y}}{x-y}&underset{mathcal{D}}{=}1,\
end{split}
end{equation}

the trigonometric identity:
begin{equation}
begin{split}
1-sin^2theta&=cos^2theta=frac{1}{2}left(1+cosleft(2thetaright)right),
end{split}
end{equation}

and the fact [3]:
begin{equation}
begin{split}
int_{0}^{frac{pi}{2}} frac{mathcal{P}}{sin^2theta-y}mathrm{d}theta=0.
end{split}
end{equation}

For $y=1$ we have:
begin{equation}
begin{split}
Nleft(1right)=&frac{1}{pi} int_{0}^{1} sqrt{x}sqrt{1-x}frac{mathcal{P}_{x-1}}{x-1}mathrm{d}x=-frac{1}{pi} int_{0}^{1} sqrt{frac{x}{1-x}}mathrm{d}x\
=&-frac{2}{pi} int_{-frac{pi}{2}}^{0}cos^2thetamathrm{d}theta\
=&-frac{1}{pi} int_{-frac{pi}{2}}^{0}left(1+cosleft(2thetaright)right)mathrm{d}theta=-frac{1}{pi} left[theta+frac{1}{2}sinleft(2thetaright)right]_{-frac{pi}{2}}^{0}=-frac{1}{2},
end{split}
end{equation}

where we used the distributional equality:
begin{equation}
begin{split}
sqrt{1-x}frac{mathcal{P}_{x-1}}{x-1}underset{mathcal{D}}{=}-frac{1}{sqrt{1-x}},
end{split}
end{equation}

we have also used substitutions $x=cos^2 theta$ (so $mathrm{d}x=left|-2costhetasinthetamathrm{d}thetaright|$) (where we have preserved the orientation of our integral so we require the modulus sign).
For $y=0$ we have:
begin{equation}
begin{split}
Nleft(0right)=&frac{1}{pi} int_{0}^{1} sqrt{x}sqrt{1-x}frac{mathcal{P}_{x}}{x}mathrm{d}x=frac{1}{pi} int_{0}^{1} sqrt{frac{1-x}{x}}mathrm{d}x=frac{2}{pi} int_{-frac{pi}{2}}^{0}sin^2thetamathrm{d}theta\
=&frac{1}{pi} int_{-frac{pi}{2}}^{0}left(1-cosleft(2thetaright)right)mathrm{d}theta=frac{1}{pi} left[theta-frac{1}{2}sinleft(2thetaright)right]_{-frac{pi}{2}}^{0}=frac{1}{2},
end{split}
end{equation}

where we used the distributional equality:
begin{equation}
begin{split}
sqrt{x}frac{mathcal{P}_{x}}{x}underset{mathcal{D}}{=}frac{1}{sqrt{x}},
end{split}
end{equation}

we have also used substitutions $x=cos^2 theta$ (so $left|mathrm{d}xright|=left|2costhetasinthetamathrm{d}thetaright|$).

Bringing it all together we find:
begin{equation}
begin{split}
Nleft(yright)=&frac{1}{2}-y,
end{split}
end{equation}

for $0le yle 1$.

Conclusion:

Since $Nleft(yright)=frac{1}{2}-y$ for $0le yle 1$ (as derived above) and:
begin{equation}
begin{split}
M_{E}left(yright)=&E Nleft(frac{y}{E}right),\
L_{left(C,Dright)}left(yright)=&left(D-Cright) Nleft(frac{y-C}{D-C}right)\
K_{left(A,Bright)}left(yright)=&left(B^2-A^2right) Nleft(frac{y^2-A^2}{B^2-A^2}right),\
I_{B}left(yright)=&B^2 Nleft(frac{y^2}{B^2}right),
end{split}
end{equation}

(as derived in the section before last) we find:
begin{equation}
begin{split}
M_{E}left(yright)=&E left(frac{1}{2}-frac{y}{E}right)=frac{E}{2}-y,\
L_{left(C,Dright)}left(yright)=&left(D-Cright) left(frac{1}{2}-frac{y-C}{D-C}right)=frac{D+C}{2}-y\
K_{left(A,Bright)}left(yright)=&left(B^2-A^2right) left(frac{1}{2}-frac{y^2-A^2}{B^2-A^2}right)= frac{B^2+A^2}{2}-y^2,\
I_{B}left(yright)=&B^2 left(frac{1}{2}-frac{y^2}{B^2}right)=frac{B^2}{2}-y^2,
end{split}
end{equation}

for $0le yle E$, $Cle yle D$, $Ale yle B$, and $0le yle B$ respectively. This completes the derivation.


Footnote [1]:
We were a bit cavalier with our principal value equating:
begin{equation}
begin{split}
frac{mathcal{P}_{x+y}}{x+y}+frac{mathcal{P}_{x-y}}{x-y}underset{mathcal{D}}{=}2xfrac{mathcal{P}_{x^2-y^2}}{x^2-y^2},
end{split}
end{equation}

so we shall verify our procedure was valid here:
begin{equation}
begin{split}
&frac{mathcal{P}_{x+y}}{x+y}+frac{mathcal{P}_{x-y}}{x-y}=lim_{varepsilon_1,varepsilon_2rightarrow 0}left[frac{x+y}{left(x+yright)^2+varepsilon_1^2}+frac{x-y}{left(x-yright)^2+varepsilon_2^2}right]\
=&lim_{varepsilon_1,varepsilon_2rightarrow 0}left[frac{left(x^2-y^2right)left(x-yright)+varepsilon_2^2left(x+yright)+left(x^2-y^2right)left(x+yright)+varepsilon_1^2left(x-yright)}{left(left(x+yright)^2+varepsilon_1^2right)left(left(x-yright)^2+varepsilon_2^2right)}right]\
=&lim_{varepsilon_1,varepsilon_2rightarrow 0}left[frac{2xleft(x^2-y^2right)+varepsilon_2^2left(x+yright)+varepsilon_1^2left(x-yright)}{left(x^2-y^2right)^2+varepsilon_1^2left(x-yright)^2+varepsilon_2^2left(x+yright)^2+varepsilon_1^2varepsilon_2^2}right]\
=&begin{cases}
frac{2x}{x^2-y^2}, &text{ for } x^2ne y^2\
lim_{varepsilon_1,varepsilon_2rightarrow 0}left[frac{left(x+yright)}{left(x+yright)^2+varepsilon_1^2}right]=frac{mathcal{P}_{x+y}}{x+y}=frac{1}{2y}, &text{ for } x= yne 0\
lim_{varepsilon_1,varepsilon_2rightarrow 0}left[frac{left(x-yright)}{left(x-yright)^2+varepsilon_2^2}right]=frac{mathcal{P}_{x-y}}{x-y}=-frac{1}{2y}, &text{ for } x= -yne 0\
lim_{varepsilon_1,varepsilon_2rightarrow 0}left[frac{2xleft(x^2+varepsilon_2^2+varepsilon_1^2right)}{x^4+left(varepsilon_1^2 +varepsilon_2^2right)x^2+varepsilon_1^2varepsilon_2^2}right]_{x=0}=0, &text{ for } x= pm y= 0\
end{cases}
end{split}
end{equation}

where we use the definition of principal value distributions above, if we compare this to:
begin{equation}
begin{split}
&2xfrac{mathcal{P}_{x^2-y^2}}{x^2-y^2}=lim_{varepsilonrightarrow 0}left[2xfrac{x^2-y^2}{left(x^2-y^2right)^2+varepsilon^2}right]\
=&begin{cases}
frac{2x}{x^2-y^2}, &text{ for } x^2ne y^2\
0, &text{ for } x= yne 0\
0, &text{ for } x= -yne 0\
lim_{varepsilonrightarrow 0}left[2xfrac{x^2}{x^4+varepsilon^2}right]_{x=0}=0, &text{ for } x= pm y= 0\
end{cases}
end{split}
end{equation}

we see that the distributions are equal everywhere except at two points $x=y$ and $x=-y$ (where $yne 0$), but here they only differ by a finite amount ($pmfrac{1}{2y}$) at a finite number (2) of points (unless $y=0$ in which case they do not differ at any points) and thus they can be considered to be distributionally equal.

Footnote [2]:
We leave proof of the distributional equality:
begin{equation}
begin{split}
Efrac{mathcal{P}_{Ew-y}}{Ew-y}underset{mathcal{D}}{=} frac{mathcal{P}_{w-frac{y}{E}}}{w-frac{y}{E}},
end{split}
end{equation}

as an exercise for the reader.

Footnote [3]:
It would be nice to prove $int_{0}^{frac{pi}{2}} frac{mathcal{P}_{sin^2theta-y}}{sin^2theta-y}mathrm{d}theta=0
$
using symmetry but we can show it in a somewhat clumsy way for $0<y<1$. We have, where we use the second part of the principle value definition (that removes the pole $y=sintheta$, i.e. $theta=phi$, from the integration range by taking explicit limits), the result:
begin{equation}
begin{split}
&int_{0}^{frac{pi}{2}} frac{mathcal{P}_{sin^2theta-y}}{sin^2theta-y}mathrm{d}theta=int_{0}^{frac{pi}{2}} frac{mathcal{P}_{sinleft(theta+phiright)sinleft(theta-phiright)}}{sinleft(theta+phiright)sinleft(theta-phiright)}mathrm{d}theta\
=&lim_{varepsilonrightarrow 0}left[int_{0}^{phi-varepsilon} frac{1}{sinleft(theta+phiright)sinleft(theta-phiright)}mathrm{d}theta+int_{phi+varepsilon}^{frac{pi}{2}} frac{1}{sinleft(theta+phiright)sinleft(theta-phiright)}mathrm{d}thetaright]\
=&lim_{varepsilonrightarrow 0}left[left[frac{1}{sin left(2phiright)}lnleft|frac{sinleft(theta-phiright)}{sinleft(theta+phiright)}right|right]_{0}^{phi-varepsilon}+left[frac{1}{sin left(2phiright)}lnleft|frac{sinleft(theta-phiright)}{sinleft(theta+phiright)}right|right]_{phi+varepsilon}^{frac{pi}{2}} right]\
=&frac{1}{sin left(2phiright)}lim_{varepsilonrightarrow 0}left[lnleft|frac{sinleft(-varepsilonright)}{sinleft(2phi-varepsilonright)}right|-lnleft|frac{sinleft(-phiright)}{sinleft(phiright)}right|right.\
&~~~~~~~~~~~~~~~~~~~~~~~~~left.+lnleft|frac{sinleft(frac{pi}{2}-phiright)}{sinleft(frac{pi}{2}+phiright)}right|-lnleft|frac{sinvarepsilon}{sinleft(2phi+varepsilonright)}right| right]\
=&frac{1}{sin left(2phiright)}lim_{varepsilonrightarrow 0}left[lnleft|frac{sinvarepsilon}{sinleft(2phiright)cosvarepsilon-cosleft(2phiright)sinvarepsilon}right|right.\
&~~~~~~~~~~~~~~~~~~~~~~~~~left.-lnleft|frac{sinvarepsilon}{sinleft(2phiright)cosvarepsilon+cosleft(2phiright)sinvarepsilon}right| right]\
=&frac{1}{sin left(2phiright)}lim_{varepsilonrightarrow 0}left[lnleft|frac{1+cotleft(2phiright)varepsilon}{1-cotleft(2phiright)varepsilon}+mathcal{O}left(varepsilon^2right)right| right]\
=&frac{1}{sin left(2phiright)}lim_{varepsilonrightarrow 0}left[lnleft|1+2cotleft(2phiright)varepsilon+mathcal{O}left(varepsilon^2right)right| right]\
=&lim_{varepsilonrightarrow 0}left[frac{2cotleft(2phiright)}{sin left(2phiright)}varepsilon+mathcal{O}left(varepsilon^2right)right]=0,
end{split}
end{equation}

where we have used the substitution $sinphi=y$ (valid since $0<y<1$, it means $0<phi<frac{pi}{2}$) and the trigonometric identities $sin^2theta-sin^2phi=sinleft(theta+phiright)sinleft(theta-phiright)$ and $sinleft(x+yright)=sin xcos y +cos xsin y$ as well as the Taylor expansions $sin x=x+mathcal{O}left(x^3right)$, $cos x=1+mathcal{O}left(x^2right)$, and $lnleft(1+xright)=x+mathcal{O}left(x^2right)$. We have also used the fact that:
begin{equation}
begin{split}
&frac{mathrm{d}}{mathrm{d}theta}left[frac{1}{sin left(2phiright)}lnleft|frac{sinleft(theta-phiright)}{sinleft(theta+phiright)}right|right]\
=&frac{1}{sin left(2phiright)}frac{cosleft(theta-phiright)sinleft(theta+phiright)-sinleft(theta-phiright)cosleft(theta+phiright)}{sinleft(theta-phiright)sinleft(theta+phiright)}\
=&frac{1}{sinleft(theta-phiright)sinleft(theta+phiright)},
end{split}
end{equation}

where we have used the trigonometric identity:
$sinleft(2thetaright)=sinleft(theta+phiright)cosleft(theta-phiright)-cosleft(theta+phiright)sinleft(theta-phiright)$. We were able to substitute the antiderivative because neither of the integrals we considered featured poles (the poles were at $theta =phi$).

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Math Genius: Computing $int_Q frac{xy}{x^2+y^2}dxdy$

Original Source Link

I’m asked to compute the following integral:
$$int_Q frac{xy}{x^2+y^2}dxdy qquad Q=[0,1]^2$$

Solution:
First I’m going to study if the integral is convergent. To do this, we notice that
$$int_Q frac{xy}{x^2+y^2}dxdy < infty iff int_S frac{xy}{x^2+y^2}dxdy<infty$$
where $S={(x,y) in mathbb{R}^2 | xgeq0, ygeq 0, x^2+y^2leq1}$ and this is clear because the difference between $int_Q f(x,y)dxdy$ and $int_S f(x,y)dxdy$ is a proper integral.

Computing we have:
$$int_S frac{xy}{x^2+y^2}dxdy=lim_{epsilon to 0}int_{epsilon}^1int_0^{frac{pi}{2}}rho^3sinthetacostheta drho dtheta=frac{1}{8}$$
and so I’m granted the convergence.

Now I have to compute the real integral, as we have assured that it’s convergent.
$$int_Q frac{xy}{x^2+y^2}dxdy=int_0^1 int_0^1 frac{xy}{x^2+y^2}dxdy= int_0^1 frac{y}{2} int_0^1 frac{2x}{x^2+y^2}dxdy = frac{1}{2} int_0^1 y(log(1+y^2)-log(y^2))dy =$$
$$ = frac{log2}{2}-frac{1}{2}lim_{epsilon to 0}int_{epsilon}^1ylog(y^2)dy= frac{log2}{2}$$

I checked the result and it’s correct, but I’m asking for a review of the process: did I do anything wrong?

As you wrote it, it looks like you claim
$$ I = frac{1}{2} int_0^1 y ln (y^2) dy = 0.$$

But this is not the case:
begin{align*}
frac{1}{2} int_0^1 y ln(y^2) dy &= frac 1 4 int_{0}^1 ln(y^2)2y dy \
&= frac 1 4 int_0^1 ln(u) du.
end{align*}

Using the fact the $u ln u – u $ is a primitive of $ln u$ you get

$$ I = frac 1 4 big(- 1 – (0 – 0)big) = frac {-1}{4}.$$

Same goes for

$$J = frac 1 2 int_0^1 y ln (1 + y^2) dy neq frac {ln(2)} {2}.$$

but rather

begin{align*}
frac{1}{2} int_0^1 y ln (1 + y^2) &= frac{1}{4} int_0^1 ln(1+y^2) 2ydy \
&= frac{1}{4} int_1^2 ln(u) du \
&= frac{1}{4} (2 ln 2 – 2 – (1cdotln 1 – 1)) \
&= frac 1 4 (2 ln 2 -1) \
&= frac{ln(2)}{2} – frac{1}{4}.
end{align*}

So the integral is
$$J – I = left(frac{ln(2)}{2} -frac{1}{4}right) – frac{-1}{4} =frac{ln(2)}{2}.$$

Concerning the first part it seems easier to me to do the following

$$ leftvert int_Q f(x,y)dxdy rightvert leq int_Q vert f(x,y)vert dxdy leq int_{D_R^+} vert f(x,y)vert dxdy$$

where $D_R^+$ is the positive par of a disc of radius $R$ with $R$ such that $Q subset D_R^+.$

Using polar coordinates you will have the following inside the integral

$$ left vert frac{r^2 cos sin}{R^2} r right vertleq r leq R$$ and
$ int_Q f(x,y) dxdy < infty.$

I’m not too impressed by the way you proved convergence. The place where the denominator of the integrand becomes $0$ is $x=0, y=0$ and $(0,0)$ is in both $Q$ and $S.$ A better way is let $$Q’={(x,y)|x^2+y^2<b^2}$$ where $b<1,$ integrate over $Qbackslash Q’$ and take the limit as $b to 0.$ Split the region $Qbackslash Q’$into two parts depending on whether $y le x$ or $y ge x.$ Here’s how to do the integral over the ‘lower’ part where $y le x$. Note that the boundary $x=1$ is $r=sectheta$ in polar coordinates. Then the integral over the lower part is, in polar coordinates, $$int_0^{pi/4}int_b^{sec theta}frac{(rcostheta)(rsin theta)}{r^2}rdrdtheta$$ which is easily integrated. The top half, where the boundary $y=1$ is $r=csc theta$ is equally easy to integrate. Add the top and bottom parts, take the limit as $b to 0$ and you’re done.

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Math Genius: Efficient/faster methods to find the general closed form of $int _0^1frac{ln left(ax^2+bright)}{x+1}:dx$

Original Source Link

I managed to find it using the good ol’ Feynman’s trick but it was an extremely long process, so i’d like to know if theres more efficient/faster methods out there that you guys could provide. thanks.

Ill share my attempt, but it is ridiculously long.

My attempt.

$$Ileft(aright)=int _0^1frac{ln left(ax^2+bright)}{x+1}:dx$$
$$I’left(aright)=int _0^1frac{x^2}{left(x+1right)left(ax^2+bright)}:dx=frac{1}{a+b}int _0^1frac{bleft(x-1right)}{ax^2+b}+frac{1}{x+1}:dx$$
$$I’left(aright)=frac{1}{a+b}left(frac{b}{2a}ln left(a+bright)-frac{b}{2a}ln left(bright)-frac{sqrt{b}}{sqrt{a}}arctan left(frac{sqrt{a}}{sqrt{b}}right)+ln left(2right)right)$$
$$int _b^aI’left(aright):da=frac{b}{2}int _b^afrac{ln left(a+bright)}{aleft(a+bright)}:da-frac{b}{2}ln left(bright)int _b^afrac{1}{aleft(a+bright)}:da-sqrt{b}underbrace{int _b^afrac{arctan left(sqrt{frac{a}{b}}right)}{sqrt{a}left(a+bright)}:da}_{u=sqrt{frac{a}{b}}}+ln left(2right)int _b^afrac{1}{a+b}:da$$
Now lets calculate $Ileft(bright)$.
$$Ileft(bright)=ln left(bright)int _0^1frac{1}{x+1}:dx+int _0^1frac{ln left(x^2+1right)}{x+1}:dx=ln left(bright)ln left(2right)+frac{3ln ^2left(2right)}{4}-frac{pi ^2}{48}$$
Now resuming on the original expression:
$$Ileft(aright)-ln left(bright)ln left(2right)-frac{3ln ^2left(2right)}{4}+frac{pi ^2}{48}=frac{1}{2}underbrace{int _b^afrac{ln left(a+bright)}{a}:da}_{a=bt}-frac{1}{2}int _b^afrac{ln left(a+bright)}{a+b}:da-frac{ln left(bright)}{2}int _b^afrac{1}{a}:da+frac{ln left(bright)}{2}int _b^afrac{1}{a+b}:da-2int _1^{sqrt{frac{a}{b}}}frac{arctan left(uright)}{u^2+1}:du+ln left(2right)ln left(a+bright)-ln left(2right)ln left(2bright)$$
$$Ileft(aright)=frac{ln left(bright)}{2}int _1^{frac{a}{b}}frac{1}{t}:dt+frac{1}{2}int _1^{frac{a}{b}}frac{ln left(t+1right)}{t}:dt-frac{ln ^2left(a+bright)}{4}+frac{ln ^2left(2bright)}{4}-frac{ln left(bright)ln left(aright)}{2}+frac{ln ^2left(bright)}{2}+frac{ln left(bright)ln left(a+bright)}{2}-frac{ln left(bright)ln left(2bright)}{2}-arctan ^2left(sqrt{frac{a}{b}}right)+frac{pi ^2}{16}+ln left(2right)ln left(a+bright)-ln left(2right)ln left(2bright)+ln left(bright)ln left(2right)+frac{3ln ^2left(2right)}{4}-frac{pi ^2}{48}$$
$$Ileft(aright)=frac{1}{2}ln left(bright)ln left(frac{a}{b}right)+frac{1}{2}underbrace{int _0^{frac{a}{b}}frac{ln left(t+1right)}{t}:dt}_{t=-t}-frac{1}{2}int _0^1frac{ln left(t+1right)}{t}:dt-frac{ln ^2left(a+bright)}{4}+frac{ln ^2left(2right)}{4}+frac{ln left(2right)ln left(bright)}{2}+frac{ln ^2left(bright)}{4}-frac{ln left(bright)ln left(aright)}{2}+frac{ln ^2left(bright)}{2}+frac{ln left(bright)ln left(a+bright)}{2}-frac{ln left(2right)ln left(bright)}{2}-frac{ln ^2left(bright)}{2}-arctan ^2left(sqrt{frac{a}{b}}right)+frac{pi ^2}{24}+ln left(2right)ln left(a+bright)-ln ^2left(2right)-ln left(2right)ln left(bright)+ln left(bright)ln left(2right)+frac{3ln ^2left(2right)}{4}$$
$$Ileft(aright)=frac{ln left(bright)ln left(aright)}{2}-frac{ln ^2left(bright)}{2}+frac{1}{2}int _0^{-frac{a}{b}}frac{ln left(1-tright)}{t}dt-frac{pi ^2}{24}-frac{ln ^2left(a+bright)}{4}+frac{ln ^2left(bright)}{4}-frac{ln left(bright)ln left(aright)}{2}+frac{ln left(bright)ln left(a+bright)}{2}-arctan ^2left(sqrt{frac{a}{b}}right)+frac{pi ^2}{24}+ln left(2right)ln left(a+bright)$$
To solve the integral remaning we can use the following identity:
$$text{Li}_2left(zright)=-int _0^zfrac{ln left(1-tright)}{t}:dt$$
After using it and simplifying a bit more we finally arrive at the solution being:
$$boxed{Ileft(aright)=-frac{ln ^2left(bright)}{4}-frac{text{Li}_2left(-frac{a}{b}right)}{2}-frac{ln ^2left(a+bright)}{4}+frac{ln left(bright)ln left(a+bright)}{2}-arctan ^2left(sqrt{frac{a}{b}}right)+ln left(2right)ln left(a+bright)}$$
One of the integrals i used can be proved with this:
$$int _0^1frac{ln left(x^2+1right)}{x+1}:dx=-frac{text{Li}_2left(-1right)}{2}-frac{ln ^2left(2right)}{4}-arctan ^2left(1right)+ln ^2left(2right)=frac{pi ^2}{24}+frac{3ln ^2left(2right)}{4}-frac{pi ^2}{16}=frac{3ln ^2left(2right)}{4}-frac{pi ^2}{48}$$

The value of both $a$ and $b$ don’t matter as long as $a,b >0$. We can split it up to get that

$$int_0^1frac{log(ax^2+b)}{x+1}:dx = log a log 2 + int_0^1frac{log(x^2+c)}{x+1}:dx$$

where we have a new parameter $c equiv frac{b}{a}$. Then taking the derivative we have that

$$I'(c) = int_0^1frac{1}{(x^2+c)(x+1)}:dx = frac{1}{c+1}int_0^1 frac{1}{x+1}-frac{x-1}{x^2+c}:dx$$

$$= frac{log 2}{c+1} – frac{log c}{2(c+1)} + frac{arctanleft(frac{1}{sqrt{c}}right)}{sqrt{c}(c+1)}$$

which can be solved in similar ways as before, but now it’s only one variable.


For slight completeness sake, continuing on a bit further we get that

$$I(a,b) = log 2 log(a+b) – arctan^2left(sqrt{frac{a}{b}}right)+int_0^1 frac{2log t}{t+1}:dx – frac{1}{2}int_0^{frac{b}{a}} frac{log t}{t+1}:dt$$

where the value of the last two integrals can be given by special functions.

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Math Genius: Computing $int_Q frac{xy}{x^2+y^2}dxdy$

Original Source Link

I’m asked to compute the following integral:
$$int_Q frac{xy}{x^2+y^2}dxdy qquad Q=[0,1]^2$$

Solution:
First I’m going to study if the integral is convergent. To do this, we notice that
$$int_Q frac{xy}{x^2+y^2}dxdy < infty iff int_S frac{xy}{x^2+y^2}dxdy<infty$$
where $S={(x,y) in mathbb{R}^2 | xgeq0, ygeq 0, x^2+y^2leq1}$ and this is clear because the difference between $int_Q f(x,y)dxdy$ and $int_S f(x,y)dxdy$ is a proper integral.

Computing we have:
$$int_S frac{xy}{x^2+y^2}dxdy=lim_{epsilon to 0}int_{epsilon}^1int_0^{frac{pi}{2}}rho^3sinthetacostheta drho dtheta=frac{1}{8}$$
and so I’m granted the convergence.

Now I have to compute the real integral, as we have assured that it’s convergent.
$$int_Q frac{xy}{x^2+y^2}dxdy=int_0^1 int_0^1 frac{xy}{x^2+y^2}dxdy= int_0^1 frac{y}{2} int_0^1 frac{2x}{x^2+y^2}dxdy = frac{1}{2} int_0^1 y(log(1+y^2)-log(y^2))dy =$$
$$ = frac{log2}{2}-frac{1}{2}lim_{epsilon to 0}int_{epsilon}^1ylog(y^2)dy= frac{log2}{2}$$

I checked the result and it’s correct, but I’m asking for a review of the process: did I do anything wrong?

As you wrote it, it looks like you claim
$$ I = frac{1}{2} int_0^1 y ln (y^2) dy = 0.$$

But this is not the case:
begin{align*}
frac{1}{2} int_0^1 y ln(y^2) dy &= frac 1 4 int_{0}^1 ln(y^2)2y dy \
&= frac 1 4 int_0^1 ln(u) du.
end{align*}

Using the fact the $u ln u – u $ is a primitive of $ln u$ you get

$$ I = frac 1 4 big(- 1 – (0 – 0)big) = frac {-1}{4}.$$

Same goes for

$$J = frac 1 2 int_0^1 y ln (1 + y^2) dy neq frac {ln(2)} {2}.$$

but rather

begin{align*}
frac{1}{2} int_0^1 y ln (1 + y^2) &= frac{1}{4} int_0^1 ln(1+y^2) 2ydy \
&= frac{1}{4} int_1^2 ln(u) du \
&= frac{1}{4} (2 ln 2 – 2 – (1cdotln 1 – 1)) \
&= frac 1 4 (2 ln 2 -1) \
&= frac{ln(2)}{2} – frac{1}{4}.
end{align*}

So the integral is
$$J – I = left(frac{ln(2)}{2} -frac{1}{4}right) – frac{-1}{4} =frac{ln(2)}{2}.$$

Concerning the first part it seems easier to me to do the following

$$ leftvert int_Q f(x,y)dxdy rightvert leq int_Q vert f(x,y)vert dxdy leq int_{D_R^+} vert f(x,y)vert dxdy$$

where $D_R^+$ is the positive par of a disc of radius $R$ with $R$ such that $Q subset D_R^+.$

Using polar coordinates you will have the following inside the integral

$$ left vert frac{r^2 cos sin}{R^2} r right vertleq r leq R$$ and
$ int_Q f(x,y) dxdy < infty.$

I’m not too impressed by the way you proved convergence. The place where the denominator of the integrand becomes $0$ is $x=0, y=0$ and $(0,0)$ is in both $Q$ and $S.$ A better way is let $$Q’={(x,y)|x^2+y^2<b^2}$$ where $b<1,$ integrate over $Qbackslash Q’$ and take the limit as $b to 0.$ Split the region $Qbackslash Q’$into two parts depending on whether $y le x$ or $y ge x.$ Here’s how to do the integral over the ‘lower’ part where $y le x$. Note that the boundary $x=1$ is $r=sectheta$ in polar coordinates. Then the integral over the lower part is, in polar coordinates, $$int_0^{pi/4}int_b^{sec theta}frac{(rcostheta)(rsin theta)}{r^2}rdrdtheta$$ which is easily integrated. The top half, where the boundary $y=1$ is $r=csc theta$ is equally easy to integrate. Add the top and bottom parts, take the limit as $b to 0$ and you’re done.

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Math Genius: One way to solve $int_{0}^{infty}3Big(frac{e^{-x^3}}{x+1}+frac{xe^{-x^3}}{x^3+1}-frac{e^{-x^3}}{x^3+1}Big)dx=G$

Original Source Link

It’s a simple question we have :

$$int_{0}^{infty}3Big(frac{e^{-x^3}}{x+1}+frac{xe^{-x^3}}{x^3+1}-frac{e^{-x^3}}{x^3+1}Big)dx=G$$

Where $G$ is the Gompertz constant

It has a simple antiderivative wich is equal to :
$$eE_i(-x^3-1)$$

Where $E_i(x)$ is the Exponential integral

Applying the Fundamental theorem of calculus we get the desired result .

My question :

Have you another way to prove it ?

Thanks in advance for all your contributions.

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Math Genius: $ lim_{ttoinfty}int_{0}^{t}|g(s)|ds=0?$, when $lim_{ttoinfty} |g(t)|=0$

Original Source Link

I was proving something with about differential equations and I stumbled on a problem which was cited in my book as trivial, but I don´t see it clearly:

So we have some function $g$ with:
$$lim_{ttoinfty} |g(t)|=0$$

How to show that:

$$ lim_{ttoinfty}int_{0}^{t}|g(s)|ds=0?$$

* EDIT: *
The question is related to the answer to the following post: Diff equations-Duhamels formula for $ttoinfty$
What was meant here by “Then it’s easy to show that the r.h.s. converges to zero.”

If $g$ (measurable) satisfies $lim_{trightarrowinfty}int^t_0|g(s)|,ds=0$ then $int^A_0|g(s)|,dsleqlim_{trightarrowinfty}int^t_0|g(s)|,ds=0$ which means that $g=0$ a.s on any interval $[0,A]$ and so over on the whole real line. Bottom line, the only solution to your problem is $gequiv0$ (a.s).

Having $ gleft(tright)underset{tto +infty}{longrightarrow}0 $, it is possible for $ int_{0}^{+infty}{left|gleft(tright)right|mathrm{d}t} $ to converge to some value other than $ 0 $, which is the case for $ g:tmapstomathrm{e}^{-t^{2}} $. It is also possible for $ int_{0}^{+infty}{left|gleft(tright)right|mathrm{d}t} $ to diverge, taking, for example, $ g:tmapstofrac{sin{t}}{t} $.

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Math Genius: $ lim_{ttoinfty}int_{0}^{t}|g(s)|ds=0?$, when $lim_{ttoinfty} |g(t)|=0$

Original Source Link

I was proving something with about differential equations and I stumbled on a problem which was cited in my book as trivial, but I don´t see it clearly:

So we have some function $g$ with:
$$lim_{ttoinfty} |g(t)|=0$$

How to show that:

$$ lim_{ttoinfty}int_{0}^{t}|g(s)|ds=0?$$

* EDIT: *
The question is related to the answer to the following post: Diff equations-Duhamels formula for $ttoinfty$
What was meant here by “Then it’s easy to show that the r.h.s. converges to zero.”

If $g$ (measurable) satisfies $lim_{trightarrowinfty}int^t_0|g(s)|,ds=0$ then $int^A_0|g(s)|,dsleqlim_{trightarrowinfty}int^t_0|g(s)|,ds=0$ which means that $g=0$ a.s on any interval $[0,A]$ and so over on the whole real line. Bottom line, the only solution to your problem is $gequiv0$ (a.s).

Having $ gleft(tright)underset{tto +infty}{longrightarrow}0 $, it is possible for $ int_{0}^{+infty}{left|gleft(tright)right|mathrm{d}t} $ to converge to some value other than $ 0 $, which is the case for $ g:tmapstomathrm{e}^{-t^{2}} $. It is also possible for $ int_{0}^{+infty}{left|gleft(tright)right|mathrm{d}t} $ to diverge, taking, for example, $ g:tmapstofrac{sin{t}}{t} $.

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Math Genius: Proving that the Kernel of an Integral Equation is Weakly Singular

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I have a simple problem of deducing whether the kernel $$k(x,t) := log |x-t|$$ is weakly singular or not. I have seen many basic examples of how to do this but I can’t make the link to this.

I know that the kernel is weakly singular if $$|k(x,t)| leq C|x-t|^{- alpha}$$

Does it have something to do with $log 0$ being undefined and $|x-t| rightarrow 0$ ? Any help would be appreicated.

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Math Genius: Divergence of definite integral

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Prove that $intlimits_0^1 frac{e^{-st}}{t^2},dt$ diverges.

My progress: I have encountered several problems so far where integral was improper and given integration was possible to express explicitly. However, my main struggles in this problem are $1)$ I can not even find to which function this is equivalent and $2)$ How can I use limit concept to prove the divergence? Any help or ideas would be strongly welcomed!

If $sgeq0$, $stleq s$ for $0<t<1$ and so, $frac{e^{-st}}{t^2}geq frac{e^{-s}}{t^2}$. From $int^1_0frac{1}{t^2},dt=infty$ you get that the integral of interest diverges to infinity. A similar argument works if $s<0$.

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