Evaluating $\int_{0}^{\infty}\frac{\sin^2(x)}{x^2(1+x^2)}\,dx$

Question

As the title says I’m wondering what is wrong with my solution process in evaluating

$$\int_{0}^{\infty}\frac{\sin^2(x)}{x^2(1+x^2)}\,dx$$

Here is what I do:

$$I=\int_{0}^{\infty}\frac{\sin^2(x)}{x^2(1+x^2)}\,dx$$

Consider $$\frac{1}{x^2}=\int_{0}^{\infty}e^{-tx^2}\,dt$$

So

$$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin^2(x)e^{-tx^2}}{1+x^2}\,dtdx$$

By Fubini’s theorem we can say

$$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin^2(x)e^{-tx^2}}{1+x^2}\,dxdt$$

Now we will evaluate the inner integral which is

$$J=\int_{0}^{\infty}\frac{\sin^2(x)e^{-tx^2}}{x^2+1}\,dx$$

Consider

$$f(z)=\frac{(e^{iz}-e^{-iz})^2e^{-tz^2}}{4(z^2+1)}$$

Consider a semi-circular contour $C$ in the upper-half of the complex plane with a radius of $R$, and the upper curve from $R$ to $-R$ being demoted as capital gamma.

So we have

$$\oint_{C}f(z)\,dz=\int_{\Gamma}f(z)\,dz+\int_{-R}^{R}f(x)\,dx$$

Calculating the residues of $f(z)$ in the contour $C$

$$\frac{1}{4}\lim_{x\to i}(z-i)\frac{(e^{iz}-e^{-iz})^2e^{-tz^2}}{(z+i)(z-i)}$$

$$=\frac{e^t(e^{-1}-e)^2}{8i}$$

And so

$$\oint_{C}f(z)\,dz=2\pi i\sum Res(f(z))= \frac{1}{4}\pi e^t(e^{-1}-e)^2$$

So this yields

$$\int_{\Gamma}f(z)\,dz+\int_{-R}^{R}f(x)\,dx=\frac{1}{4}\pi e^t(e^{-1}-e)^2$$

And I’m the limit as R goes to infinity the Gamma integral goes to zero by Jordan’s lemma. And by symmetry of even functions we have

$$J=\int_{0}^{\infty}\frac{\sin^2(x)e^{-tx^2}}{x^2+1}\,dx=\frac{\pi}{8} e^t(e^{-1}-e)^2$$

Finally this means that

$$I=\frac{\pi}{8}(e^{-1}-e)^2\int_{0}^{\infty}e^t\,dt$$

But this is divergent.

Where did I go wrong?

Answer 1

Hint: if you want to evaluate the integral start by proving that: $$\int_0^\infty \cos (ax) e^{-ux^2} dx = \frac12 e^{-\frac{a^2}{2u}}$$

Answer 2

$$f(z)=\frac{(e^{iz}-e^{-iz})^2e^{-tz^2}}{4(z^2+1)}$$

Consider a semi-circular contour $C$ in the upper-half of the complex plane with a radius of $R$, and the upper curve from $R$ to $-R$ being demoted as capital gamma.

The mistake is from here. If you choose the semi-circle on the upper half plane with radius $R$, the integrand $f(z)$ is divergent when the imaginary part approaches to infinity. For example, if we test the point close to the imaginary axis, i.e: $z\approx iR$, the numerator gives

$$(e^{iz}-e^{-iz})^2e^{-tz^2}\sim e^{2R}e^{tR^2}\longrightarrow\infty, ~~~\text{as}~~R\to\infty$$

Also note that you cannot choose the lower half plane either, since the term $$e^{-tz^2}=e^{tR^2}\longrightarrow\infty$$

no matter it is close to $z=iR$ or close to $z=-i R$

Answer 3

Since the error in your proof has already been pointed out, here are a couple ways to approach this integral:

Feynman's trick: Inserting a parameter appropriately is the name of the game; define the integral

$$I(a)=\int_0^\infty\frac{\sin^2 ax}{x^2(x^2+b^2)}dx$$

Differentiate twice with respect to $a$

$$I''(a)=\int_{-\infty}^\infty\frac{\cos 2ax}{b^2+x^2}dx=\Re \int_{-\infty}^\infty\frac{e^{2iax}}{1+x^2}dx$$

The last integral can be done using the usual combination of residue calculus + Jordan's lemma. We obtain

$$I''(a)=\frac{\pi}{|b|} e^{-2|ab|}~~,~~ I(0)=I'(0)=0$$

which can be easily integrated for the final result.

Complex analysis only: The double zero of the denominator is an obstacle to directly applying Jordan's lemma, even though $\sin^2 x$ is expressible in terms of complex exponentials. To mitigate the issue we introduce a parameter that moves the poles away from zero so that the Jordan integral converges. We consider the convergent integral

$$J(y)=\int_{-\infty}^\infty\frac{e^{iyx}}{(x^2+\epsilon^2)(x^2+b^2)}dx$$

We note here that we can express

$$I(a)=\lim_{\epsilon\to 0^+}\left[\frac{J(0)}{4}-\frac{J(2a)+J(-2a)}{8}\right]$$

We also observe that the limit exists, despite the fact that individually each integral diverges as $\epsilon \to 0^+$. We can calculate $J$ directly by another application of Jordan's lemma + residues:

$$J(y)=\frac{\pi}{b^2-\epsilon^2}\left(\frac{e^{-\epsilon |y|}}{\epsilon}-\frac{e^{-|b| |y|}}{|b|}\right)$$

Performing the limit yields

$$I(a)=\pi\frac{ e^{-2|ab|}+2|ab|-1}{4|b|^3}$$

Answer 4

Alternatively, simplify the integral by rewriting it as \begin{align} &\int_{0}^{\infty}\frac{\sin^2x}{x^2(1+x^2)}\,dx\\ =& \int_{0}^{\infty}\left( \frac{\sin^2x}{x^2}- \frac{1}{2(1+x^2)} +\frac{\cos 2x}{2(1+x^2)}\right)\,dx\\ =&\ \frac\pi2-\frac\pi4+\frac\pi {4e^2} \end{align}