$\int _0^{\pi }\:\sum _{n=0}^{\infty \:}\frac{n\cdot \sin \left(nx\right)}{e^n}dx=\frac{2e}{e^2-1}$

Question

It is asked to prove: $\int _0^{\pi }\:\sum _{n=0}^{\infty \:}\frac{n\cdot \sin \left(nx\right)}{e^n}dx=\frac{2e}{e^2-1}$ I have tried to search for convergence and it gave me 0 so i can't solve it. Can you help me?

Answer 1

$$\int_0^{\pi} \sin(nx)=\frac{1}{n}\left(-\cos(nx)\right|_0^{\pi} =\frac{1}{n}\left(1-(-1)^n\right)$$ Hence, $$\int_0^{\pi} \sum_{n=0}^{\infty} \frac{n\sin(nx)}{e^n}=\sum_{n=0}^{\infty} \frac{1-(-1)^n}{e^n}=\left(\sum_{n=0}^{\infty}\frac{1}{e^n}\right)-\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{e^n}\right)$$ $$=\frac{1}{1-\left(\frac{1}{e}\right)}-\frac{1}{1-\left(-\frac{1}{e}\right)}=\frac{e}{e-1}-\frac{e}{e+1}=\boxed{\dfrac{2e}{e^2-1}}$$

Answer 2

Hint

You could use

$$\int _0^{\pi }\:\sum _{n=0}^{\infty \:}\frac{n\cdot \sin \left(nx\right)}{e^n}dx=\sum _{n=0}^{\infty \:}\int _0^{\pi }\:\frac{n\cdot \sin \left(nx\right)}{e^n}dx=\sum _{n=0}^{\infty \:}e^{-n} (1-\cos (\pi n))$$

At this point, I stop since appeared Pranav Arora's very nice answer.