how to calculate $\int_{-\pi}^{\pi} |\sin(x)|e^{-inx} \mathrm{d}x$

Question

i have a calculating question i want to calculate this integral $$\int_{-\pi}^{\pi} |\sin(x)|e^{-inx} \mathrm{d}x$$ i tried to use integration by parts but it just made it more complicated. i already have the answer (USING MAPLE)

but i would appreciate if someone could take the time to explain the steps to me.

Answer 1

Hint:$$\int_{-\pi}^\pi|\sin(x)|e^{-nix}\,\mathrm d x=2\int_0^\pi \sin(x)e^{-inx}\,\mathrm dx.$$

Use the fact that $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$, and conclude.

Answer 2

Hints: $e^{inx} =\cos (nx)+i \sin (nx)$. The integral of $|\sin x| \sin (nx)$ is $0$ because the function is odd. Also, $$\int_{-\pi}^{\pi} |\sin x| \cos(nx )\mathrm{d}x =2 \int_{0}^{\pi} |\sin x| \cos(nx )\mathrm{d}x$$ Note that absolute value sign can now be dropped. To evaluate this use the formula $$\sin A \cos B=\frac 1 2\left( \sin (A+B)+\sin (A-B)\right)$$

Answer 3

By parts:

We first need to get rid of the absolute value because it is not differentiable. As noted by others, exploiting parity we have

$$\int_{-\pi}^\pi|\sin x|e^{-inx}dx=2\int_0^\pi\sin x\,e^{-inx}dx.$$

Then

$$I:=\int_0^\pi\sin x\,e^{-inx}dx=\frac{i}n\left.\sin x\,e^{-inx}\right|_0^\pi-\frac{i}n\int_0^\pi\cos x\,e^{-inx}dx \\=-\frac{i}n\int_0^\pi\cos x\,e^{-inx}dx$$

and

$$\int_0^\pi\cos x\,e^{-inx}dx =-\frac{i}n\left.\cos x\,e^{-inx}\right|_0^\pi+\frac{i}n\int_0^\pi\sin x\,e^{-inx}dx \\=-i\frac{e^{-in\pi}+1}n+\frac{i}nI.$$

Finally,

$$\left(1-\frac1{n^2}\right)I=-\frac{e^{-in\pi}+1}{n^2}.$$

[Typos not excluded.]