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


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


It follows that


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?

  • $\begingroup$ Next time simply put dollar symbols around your mathematical expression for them to render correctly. $\endgroup$
    – Digitallis
    Jun 11, 2023 at 20:36
  • $\begingroup$ That is why you asked yesterday math.stackexchange.com/questions/4716356/… $\endgroup$
    – MathFail
    Jun 11, 2023 at 20:51
  • 1
    $\begingroup$ This is the same author, who asked yesterday and want to solve the integral in this OP. @AnneBauval $\endgroup$
    – MathFail
    Jun 11, 2023 at 21:08
  • 2
    $\begingroup$ @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. $\endgroup$ Jun 11, 2023 at 21:10
  • 2
    $\begingroup$ 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. $\endgroup$
    – Person
    Jun 11, 2023 at 21:51

1 Answer 1


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.

  • $\begingroup$ Lai, what is the proof of Lobachevsky integral formula ? Could you write it or put a link containing the proof ? $\endgroup$
    – Angelo
    Jun 12, 2023 at 9:10
  • $\begingroup$ You may read this article hal.science/hal-01539895v2/document. Wish it helps. $\endgroup$
    – Lai
    Jun 12, 2023 at 10:47
  • $\begingroup$ Lai, it seems to be the same proof that the OP wrote in his question. Am I wrong? $\endgroup$
    – Angelo
    Jun 12, 2023 at 11:43
  • $\begingroup$ OP’s solution is more detailed. I am lazy and just use a famous trick. $\endgroup$
    – Lai
    Jun 12, 2023 at 12:24
  • $\begingroup$ Lai, I was speaking about the proof of Lobachevsky formula in that link you gave me, that proof seems to be the same one that the OP wrote in his question. Do you agree with me or am I wrong? $\endgroup$
    – Angelo
    Jun 12, 2023 at 12:44

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