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Given the integral $(1)$

$$\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot e^{-x^2}\sin^2(x^2)=\color{red}{\sqrt{\pi}\left(\sqrt{\phi}-1\right)}\tag1$$

How does one prove $(1)$?

An attempt:

$u=x^2$ $(1)$ becomes

$${1\over 2}\int_{-\infty}^{+\infty}{\mathrm du\over u^{3/2}}\cdot e^{-u}\sin^2 u\tag2$$

Recall series $(3)$

$$e^{-x}\sin x=\sum_{n=1}^{\infty}{2^{n/2}(-x)^n\sin(n\pi/4)\over n!}\tag3$$ then $(2)$ becomes

$$\sum_{n=1}^{\infty}(-1)^n{2^{n/2}\sin(n\pi/4)\over n!}\color{blue}{\int_{-\infty}^{+\infty}u^{n-3/2}\sin u\mathrm du}\tag4$$

The blue part diverges, so how else do we tackle $(1)?$

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  • $\begingroup$ I have another result $\endgroup$ – Dr. Sonnhard Graubner Apr 1 '17 at 7:10
  • $\begingroup$ I think the OP result is correct. $\endgroup$ – Mathxx Apr 1 '17 at 7:12
  • $\begingroup$ You have different form but same result @Dr.Sonnhard $\endgroup$ – gymbvghjkgkjkhgfkl Apr 1 '17 at 7:12
  • $\begingroup$ The integral in (2) diverges, does it not? (I suppose you want $|u|$, not $u$). $\endgroup$ – uniquesolution Apr 1 '17 at 7:13
  • $\begingroup$ in (2), the interval being integrated should be $(0,\infty)$ instead of the whole $\Bbb R$ $\endgroup$ – Nick Apr 1 '17 at 7:18
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Hint. From the standard gaussian evaluation, $$ \int_{-\infty}^{+\infty}e^{-tx^2}dx=\frac{\sqrt{\pi }}{\sqrt{t}},\qquad \text{Re}\:t>0, $$one gets, by linearizing the integrand and taking the real part, $$ \int_{-\infty}^{+\infty}e^{-tx^2}\sin^2(x^2)\:dx=- \sqrt{\pi } \left(\text{Re}\:\frac{1}{2\sqrt{t+2 i}}-\frac{1}{2\sqrt{t}}\right),\qquad \text{Re}\:t>0, $$ then by integrating with respect to $t$, we obtain $$ \int_{-\infty}^{+\infty}e^{-tx^2}\frac{\sin^2(x^2)}{x^2}\:dx=\sqrt{\pi } \left(\text{Re}\:\sqrt{t+2 i}-\sqrt{t}\right),\qquad \text{Re}\:t>0, $$ and putting $t=1$ gives the announced result.

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  • $\begingroup$ This is beautiful ! I was embarked in a so big monster that I gave up. Thanks for posting this elagant answer. $\endgroup$ – Claude Leibovici Apr 1 '17 at 8:22
  • $\begingroup$ @ClaudeLeibovici You are very welcome. Thank you! $\endgroup$ – Olivier Oloa Apr 1 '17 at 11:04

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