# Evaluation of Integral $\int_0^{+\infty}\frac{\sin^2(\tan\,\!x)}{x^2}\mathrm{d}x$

Evaluation of $$\int_0^{+\infty}\dfrac{\sin^2(\tan\,\!x)}{x^2}\mathrm{d}x$$

Evaluate $\int_0^{\infty} {\sin(\tan(x)) \over x}dx$

I try to give the result through the method in the link, but I don't know what to do next.

\begin{align*} I&\overset{\mathrm{def}}{=} \int_0^{+\infty}\frac{\sin^2(\tan\,\!x)}{x^2}\,\mathrm{d}x\\ &= \frac12 \int_{-\infty}^{+\infty}\frac{\sin^2(\tan\,\!x)}{x^2}\,\mathrm{d}x\\ &=\frac12 \left(\sum_{n=-\infty}^\infty \int_{(n-\frac12)\pi}^{(n+\frac12)\pi}\right)\frac{\sin^2(\tan\,\!x)}{x^2}\,\mathrm{d}x \end{align*}

• Try to get rid of the parenthesis as you have the same integrand you started with. The next step is to substitute so that the n are in the denominator of $\frac 1x$ and not in the integral bounds. This then gives the expansion for tangent x. Remember that the bounds are x=$\pi\left(n\pm \frac12\right)$. Here is a graph for the apparent expansion. Commented Jun 4, 2021 at 12:23

According to Lobachevsky Integral: \begin{aligned} &\int_0^{\infty}\frac{\sin^2(\tan x)}{x^2}\mathrm{d}x\\ =&\int_0^{\infty}\frac{\sin^2(\tan x)}{\sin^2x}\cdot \frac{\sin^2 x}{x^2}\mathrm{d}x\\ =&\int_0^{\frac{\pi}{2}}\frac{\sin^2(\tan x)}{\sin^2 x}\mathrm{d}x\\ =&\int_0^{\frac{\pi}{2}}\frac{\sin^2(\tan x)}{\tan^2 x}\mathrm{d}(\tan x)\\ =&\int_0^{\infty}\frac{\sin^2 u}{u^2}\mathrm{d}u\\ =&\frac{\pi}{2} \end{aligned}

• I think it would improve your answer if you added a link to the Lobachevsky Integral Formula and maybe even described it a bit.
– robjohn
Commented Jun 5, 2021 at 20:30

The approach in the question seems to be attempting to follow the approach in my answer to the cited question, so I will adapt that approach here.

Real Manipulations \begin{align} \int_0^\infty\frac{\sin^2(\tan(x))}{x^2}\,\mathrm{d}x &=\frac12\int_{-\infty}^\infty\frac{\sin^2(\tan(x))}{x^2}\,\mathrm{d}x\tag1\\ &=\frac12\sum_{k\in\mathbb{Z}}\int_{-\pi/2}^{\pi/2}\frac{\sin^2(\tan(x))}{(x+k\pi)^2}\,\mathrm{d}x\tag2\\ &=\frac12\int_{-\pi/2}^{\pi/2}\frac{\sin^2(\tan(x))}{\sin^2(x)}\,\mathrm{d}x\tag3\\ &=\frac12\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x\tag4\\[6pt] &=\frac\pi2\tag5 \end{align} Explanation:
$$(1)$$: use the evenness of the integrand
$$(2)$$: use the periodicity of $$\tan(x)$$
$$(3)$$: substitute $$x\mapsto x/\pi$$ in $$(7)$$ from this answer and
$$\phantom{\text{(3):}}$$ take a derivative to get $$\csc^2(x)=\sum\limits_{k\in\mathbb{Z}}\frac1{(x+k\pi)^2}$$
$$(4)$$: substitute $$x\mapsto\tan^{-1}(x)$$
$$(5)$$: apply this answer

You can apply integration by parts and you get

$$\int_0^{\infty}\frac{\sin^2(\tan x)}{x^2}dx=\left.\frac{-\sin^2(\tan x)}{x}\right|_0^\infty+\int_0^{\infty}\frac{\sin(2\,\tan x)}{x}\frac{1}{\cos^2(x)}dx=$$ $$=\int_0^{\infty}\frac{\sin(2\,\tan x)}{x}\frac{1}{\cos^2(x)}dx.$$

Now, you can follow the same procedure that in Evaluate $\int_0^{\infty} {\sin(\tan(x)) \over x}dx$ to arrive to

$$\frac12\int_{-\frac12\pi}^{\frac12\pi}\frac{\sin (2\tan x)}{\tan x}\frac{1}{\cos^2(x)} dx$$

that performing the change of variable $$\tan x\to u$$

$$\frac12\int_{-\frac12\pi}^{\frac12\pi}\frac{\sin (2\tan x)}{\tan x}\frac{1}{\cos^2(x)} dx=\int_{-\infty}^{\infty}\frac{\sin (2u)}{2u} du=\int_{0}^{\infty}\frac{\sin (u)}{u} du=\frac{\pi}{2}$$

A different approach:

Consider the function $$f(z) = \frac{e^{i \alpha \tan z}}{z^{2}+\beta^{2}} , \quad (\alpha, \beta >0),$$ which has simple poles at $$z= \pm i \beta$$ and essential singularities at the half-integer multiples of $$\pi$$.

The magnitude of $$e^{i \alpha \tan z}$$ is $$\exp \left(\frac{-\alpha \sinh\left(2 \Im (z)\right)}{\cos\left(2 \Re (z)\right)+ \cosh \left(2 \Im (z)\right)} \right).$$

In the upper half plane, this value never exceeds $$1$$, which includes near the essential singularities on the real axis.

So by integrating $$f(z)$$ around a contour consisting of the real axis (with vanishingly small indentations about the half-integer multiples of $$\pi$$) and the large semicircle above it, we can conclude that $$\operatorname{PV}\int_{-\infty}^{\infty} \frac{e^{i \alpha \tan x}}{x^{2}+ \beta^{2}} \, \mathrm dx = 2 \pi i \operatorname*{Res}_{z=i \beta} f(z) = \frac{\pi}{\beta } \, e^{- \alpha \tanh \beta}.$$

Equating the real parts on both sides of the above equation, we have $$\int_{-\infty}^{\infty} \frac{\cos(\alpha \tan x)}{x^{2}+\beta^{2}} \, \mathrm dx = \frac{\pi}{\beta } \, e^{- \alpha \tanh \beta}.$$

Therefore, \begin{align} \int_{0}^{\infty} \frac{\sin^{2}(\alpha \tan x)}{x^{2}} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{2}(\alpha \tan x)}{x^{2}} \, \mathrm dx \\ &= \frac{1}{2} \lim_{\beta \to 0^{+}} \int_{-\infty}^{\infty} \frac{\sin^{2}(\alpha \tan x)}{x^{2}+\beta^{2}} \, \mathrm dx \\ &= \frac{1}{4} \lim_{\beta \to 0^{+}} \int_{-\infty}^{\infty} \frac{1- \cos (2 \alpha \tan x)}{x^{2}+\beta^{2}} \, \mathrm dx \\ &= \frac{\pi }{4} \lim_{\beta \to 0^{+}} \frac{1- e^{- 2 \alpha \tanh \beta}}{\beta} = \\ &= \frac{\pi}{4} \lim_{\beta \to 0^{+}} \frac{2 \alpha \operatorname{sech}^{2}(\beta) \, e^{-2 \alpha \tanh \beta}}{1} \\ &= \frac{\pi \alpha}{2}. \end{align}