The integral $\int_0^\infty \dfrac{x \sin(x)}{x^2+1} dx$

Question

How to calculate: $$\int_0^\infty \frac{x \sin(x)}{x^2+1} dx$$ I thought I should find the integral on the path $[-R,R] \cup \{Re^{i \phi} : 0 \leq \phi \leq \pi\}$.

I can easily take the residue in $i$ $$ Res_{z=i} \frac{x \sin(x)}{x^2+1} \quad = \quad \frac{i (e^{ii}-e^{-ii})}{2i} \quad = \quad \frac{i(\frac1e-e)}{2i} \quad = \quad \frac1{2e}-\frac e2 $$ I know that the integral on $[-R,R]$ will get nearby the real value of $ \frac12\int_0^\infty \frac{x \sin(x)}{x^2+1} dx$. That means that I'm done if I show that the integral on the other part of the path will be small, here is an attempt to do so:

$$ \left|\frac{x \sin(x)}{x^2+1} \right| \quad \leq \quad R \cdot \frac{|\sin x|}{R^2-1} $$ I could rewrite $\sin x$ but it becomes nasty. Can you help me with this?

Answer 1

Remember that $\sin x=(\exp(ix)-\exp(-ix))/2i$, and $\exp(-ix)$ grows large in the upper half plane. You have to integrate the $\exp(ix)$ part in the upper half plane, and the $\exp(-ix)$ in the lower half plane.

Also, the upper half-plane is $0<\phi<\pi$, not $2\pi$.

The trick of completing the loop does two things.
Firstly, the integral around a loop equals the sum of the residues, which are easy to calculate.

Secondly, the extra semicircle involves large values of $x$, so the function is very small along the semicircle, and the contribution along the semicircle is negligible. So, put together, the integral you want - along the diameter - plus the negligible amount along the semicircle, equals the sum of the residues.

So we make some effort to ensure the integral along the semicircle is negligible. $\exp(iz)=\exp(ix-y)$, so its absolute value is $e^{-y}$. The upper half plane is fine to integrate $\exp(iz)z/(1+z^2)$.

But $\exp(-iz)$ has absolute value $e^y$, so contributions will be large in the upper half plane. We integrate $\exp(-iz)z/(1+z^2)$ in the lower half plane instead. The integral along the diameter is the same it always was, but the residue is now anticlockwise around $-i$.

Answer 2

I will share my work with you and hope that you can tell me if I understood everything well. I used two paths:

$$ \left\{ \begin{array}{ll} A \quad = \quad [-R,R] \ \cup \ \{Re^{i\phi} \ : \ 0 \leq \phi \leq \pi\} \\ B \quad = \quad[-R,R] \ \cup \ \{Re^{i\phi} \ : \ \pi \leq \phi \leq 2\pi\} \end{array} \right. $$ I call the upper part of the circle $U$, end the lower part of the circle $L$. Both $A$ and $B$ will be followed counter clockwiseley. I split up the integral as $$ -\frac{1}{2}i \left( \int_\mathbb{R} \frac{xe^{i \phi}}{x^2+1}dx \quad - \quad \int_\mathbb{R} \frac{xe^{i \phi}}{x^2+1}dx \right) $$ Next, I took $$ \int_A \frac{xe^{i \phi}}{x^2+1}dx \quad = \quad \int_{-R}^R \frac{xe^{i \phi}}{x^2+1}dx \ + \ \int_U \frac{xe^{i \phi}}{x^2+1}dx $$ On $U$, we have, for high values of $R$ $$ \pi R \cdot \left| \frac{xe^{i \phi}}{x^2+1} \right| \ \leq \ \frac{R}{R^2-1} \cdot \frac{\pi R}{| e^{R \sin \phi}|} \ \rightarrow \ 0 $$ because $\sin \phi < 0$ on this part. It's easy to show that the integral on $L$ goes to $0$ as well, so I skip it.

It can easily be shown that the residue in $i$ equals $\frac{1}{2e}-\frac e2$, and that the residue in $-i$ equals $-\frac{1}{2e}+\frac e2$. Now we find: $$ \int_\mathbb{R}\frac{x \sin x}{x^2+1}dx \quad = \quad 2 \int_{\mathbb{R}^+}\frac{x \sin x}{x^2+1}dx \quad = \quad 2(\frac{1}{2e}-\frac e2) \quad = \quad \frac{1}{e}- e $$ I hope that someone could check this.