Question

Calc help!

ta (16 x sinly dxdy) Il the 000 Coth - 8 Wa 2 nel (94+)? +(74")* +(-98–3t) th

coth (Pi)

its three sperate equations, stuck in the first one and the last one. the middle one is 1.6 because of the infinite N

Answer 1

$\text{Let us evaluate the integral first}\\ \text{Suppose }I=16\int_0^\pi\int_0^1 x^2\sin y dxdy\\ \text{ Since the limits are constant, hence splitting them into two single integration}\\ \implies I=16\int_0^\pi \sin y dy\times \int_0^1x^2dx\\ \implies I=16\left ( -\cos y \right )_0^\pi\times \left ( \frac{x^3}{3} \right )_0^1\\ \implies I=16\times 2\times \frac{1}{3}=\color{red}\frac{32}{3}$

Now evaluating the sum of a more general infinite series as follows:

We can solve a more general sum, ~ TT 1 n2 + a2 -coth(ta) a Note that this sum satisfies the conditions in the above link. The poles lie at z = ia and z = -ia, so n=- 1 n2 + a2 cot(iz) - Res , ia 22 + a2' cot (112) -ia 22 + a2' + Res :)] Computing the residues: Res cot(12) ia 22 + a2' (z – ia) cot(12) lim ztia (z – ia)(2+ia) cot(tia 2ia = and 2) = 2 cot(772) Res -ia 22 + a2 (z +ia) cot(az) lim z+-ia (z+ia)(z – ia) cotina) 2ία Therefore, summing these we get i cotina) 1 n2 + a2 ia a coth(ra) a

In your case, you have a=1. Put a=1 in the last result;

$\sum_{n=-\infty}^\infty \frac{1}{1+n^2}=\color{red}\pi\cot h\pi$

Hence the first problem reduces to :

$16\int_0^\pi\int_0^1x^2\sin y\ dxdy\left [\frac{\sum_{n=-\infty}^\infty \frac{1}{1+n^2}}{\cot h\pi}\right ]=\color{red}\frac{32}{3}\times\frac{\pi\cot h\pi}{\cot h\pi}=\frac{32\pi}{3}$