# Evaluating $\int_{-\infty}^{\infty}\frac{\arctan(\sin^2(x))}{x^2}\,dx$. My answer: $\pi\sqrt{2(\sqrt{2}-1)}$

I think I have evaluated the following integral to be:

$$I=\int_{-\infty}^{\infty}\frac{\arctan(\sin^2(x))}{x^2}\,dx=\pi\sqrt{2(\sqrt{2}-1)}$$ I want to know if my answer is correct and if not was my method wrong? Here is what I did.

I considered rewriting the integral over the whole real line as a sum of integrals that together span the real line. More specifically I did.

$$I= \int_{-\infty}^{\infty}\frac{\arctan(\sin^2(x))}{x^2}\,dx=\sum_{n\in\mathbb{Z}}\int_{(2n-1)\frac{\pi}{2}}^{(2n+1)\frac{\pi}{2}}\frac{\arctan(\sin^2(x))}{x^2}\,dx$$ Then, considering we converge over this interval we make the substitution

$$x\longrightarrow{x+n\pi}$$

Which yields:

$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan(\sin^2(x))\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}\,dx$$ And since

$$\cot(x)=\sum_{n\in\mathbb{Z}}\frac{1}{x+n\pi}$$

It follows that

$$\csc^2(x)=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}$$

So we finally get that

$$I=2\int_{0}^{\frac{\pi}{2}}\frac{\arctan(\sin^2(x))}{\sin^2(x)}\,dx=\pi\sqrt{2(\sqrt{2}-1)}$$ Any thoughts?

• Next time simply put dollar symbols around your mathematical expression for them to render correctly. Jun 11, 2023 at 20:36
• That is why you asked yesterday math.stackexchange.com/questions/4716356/… Jun 11, 2023 at 20:51
• This is the same author, who asked yesterday and want to solve the integral in this OP. @AnneBauval Jun 11, 2023 at 21:08
• @AnneBauval Look at the denominator. Sure, the solution to this requires the solution to the link, but that doesn't mean it is a duplicate. Jun 11, 2023 at 21:10
• To be clear, I’m not trying to duplicate a post, I merely want to see if my solution process especially the part with the series representation of the integral is correct or not from people more knowledgeable than me on the subject. I also posted this to see if anyone had any other methods of solving this integral as well. Jun 11, 2023 at 21:51

Noting that the integrand can be expressed as

$$\qquad\qquad I=\int_{-\infty}^{\infty} \frac{\arctan \left(\sin ^2(x)\right)}{x^2} d x= \int_{-\infty}^{\infty} \frac{\arctan \left(\sin ^2 x\right)}{\sin ^2 x} \cdot\left(\frac{\sin x}{x}\right)^2 d x$$ and satisfying

$$\qquad\qquad \frac{\arctan \left(\sin ^2(x)\right)}{\sin^2x} = \frac{\arctan \left(\sin ^2(\pi+x)\right)}{\sin^2(\pi+x)} = \frac{\arctan \left(\sin ^2(\pi-x)\right)}{\sin^2(\pi-x)} , \frac{\arctan \left(\sin ^2(-x)\right)}{\sin^2 (-x)} = \frac{\arctan \left(\sin ^2(x)\right)}{\sin^2x}$$

Using Lobachevsky_integral_formula, we have $$I= \int_{-\infty}^{\infty} \frac{\arctan \left(\sin ^2(x)\right)}{x^2} d x= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\arctan \left(\sin ^2 x\right)}{\sin ^2 x}dx= \pi \sqrt{2(\sqrt{2}-1)}$$ where the last answer can be referred to the post.

• Lai, what is the proof of Lobachevsky integral formula ? Could you write it or put a link containing the proof ? Jun 12, 2023 at 9:10