Calculate $\int_{-\pi}^{\pi} \frac{xe^{ix}} {1+\cos^2 {x}} dx$

Question

So I'm trying to calculate $$ \int_{-\pi}^{\pi} \frac{xe^{ix}} {1+\cos^2 {x}} dx $$ knowing that if $f(a+b-x)=f(x)$ then $$ \int_{a}^{b} xf(x)dx=\frac{a+b}{2} \int_{a}^{b} f(x)dx, $$ but it doesn't apply to $f(x) = \frac{e^{ix}}{1+\cos^2 x}$ so I tried separating the function and then using $t=\pi-x$ which does not work either because I still have that complex exponential $e^{i(a+b-x)}$ which isn't equal to $e^{ix}$... Could you give me a hint ?

Answer 1

Hints/Ideas: $xe^{ix} = x\cos x + ix\sin x$, and you integrate on an interval symmetric around $0$.

1. The function $x\mapsto \frac{x\cos x}{1+\cos^2 x}$ is odd, and the function $x\mapsto \frac{x\sin x}{1+\cos^2 x}$ is even. $$ \int_{-\pi}^\pi f(x) dx = i\int_{-\pi}^\pi dx\frac{x\sin x}{1+\cos^2 x} = 2i\int_{0}^\pi dx\frac{x\sin x}{1+\cos^2 x} $$

2. Now, integrate this by integration by parts, noticing that $$ \arctan'(x) = \frac{1}{1+x^2} $$ and therefore that $$ \frac{d}{dx} \arctan\cos x = \frac{-2\sin x}{1+\cos^2 x}. $$

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Spoiler. (Details of step 2.)

Let $g=-\arctan \cos$. $$\begin{align} 2\int_{0}^\pi dx\frac{x\sin x}{1+\cos^2 x} &= \int_{0}^\pi x g'(x) dx \stackrel{\rm(IPP)}{=} \left[xg(x)\right]^\pi_0 - \int_{0}^\pi g(x) dx \\& = -\pi\arctan(-1) - \int_{0}^\pi g(x) dx = \pi\arctan(1) - \underbrace{\int_{0}^\pi g(x) dx}_{=0} = \frac{\pi^2}{4} \end{align}$$ where we used the fact that $$ \int_{0}^\pi g(x) dx = - \int_{0}^\pi \arctan \cos(x) dx = -\int_{-1}^{1} \frac{\arctan u}{\sqrt{1-u^2}} du = 0 $$ with the change of variables $u=\cos x$, and the fact that the integrand of the last integral is an odd function.

Answer 2

This can be written as $$\int_{-\pi}^{\pi}\frac{x(\cos x+i\sin x)dx}{1+\cos^2 x}$$ $$=\int_{-\pi}^{\pi}\frac{x\cos xdx}{1+\cos^2 x}+i\int_{-\pi}^{\pi}\frac{x\sin xdx}{1+\cos^2 x}$$ The first integral evaluates to $0$ (Odd function)

Whereas the second can be written as $$2i\int_{0}^{\pi}\frac{x\sin xdx}{1+\cos^2 x} \space\space\text{(even function)}$$

Next, let $$I=\int_{0}^{\pi}\frac{x\sin xdx}{1+\cos^2 x}$$replace $x\rightarrow \pi-x$ to get $$I=\int_{0}^{\pi}\frac{(\pi-x)\sin xdx}{1+\cos^2 x}$$ And add the two, to get $$I=\frac{\pi}{2}\int_{0}^{\pi}\frac{\sin xdx}{1+\cos^2 x}$$ Now take $\cos x=t$ and you're done.