# Math Genius: Finding \$left | int_{C} frac{sin{(1/z)}} {z^{2}} dz right| \$ over straight line from \$z_{1} = i\$ till \$z_{2} = 2/ pi \$

I need to find $$left | int_{C} frac{sin{(1/z)}} {z^{2}} dz right|$$ where $$C$$ is the straight line from $$z_{1} = i$$ till $$z_{2} = 2/ pi$$.

Now I would very much like some guidance with this. I see we have a singularity at $$z_{0} =0$$, but since this is not on the contour nor enclosed by the contour, I would argue this is not a problem. Also, since $$C$$ is a straight line I reckon we should use parametrization (not sure though). The line segment can be paramtrized by:
$$z(t) = i – frac{pi}{2}it,$$
with endpoints $$t=0$$ and $$t=frac{2}{pi}$$

The if we let $$f(z) = frac{sin{(1/z)}} {z^{2}}$$ and we use the fact that $$int_{C} frac{sin{(1/z)}} {z^{2}} dz = int_{t_{start}}^{t_{end}} f(z(t)) z'(t) dt$$ we find

$$int_{C} frac{sin{(1/z)}} {z^{2}} dz = int_{C} frac{sin{(1/(i – frac{pi}{2}it))}} {(i – frac{pi}{2}it)^{2}} dz,$$

but this to me makes the integral only more troublesome. So I reckon I took a wrong turn somewhere.

The function has an explicit antiderivative

$$left|int_C frac{sinleft(frac{1}{z}right)}{z^2}:dzright| = int_{frac{2}{pi}}^1 frac{sinleft(frac{1}{x}right)}{x^2}:dx = cosleft(frac{1}{x}right)Biggr|_{frac{2}{pi}}^1 = cos 1$$

$$mathbf{EDIT}$$: Sine $$z_1$$ was really $$i$$, the answer becomes

$$|cos(-i)| = cosh 1$$

Tagged : /