# How do we show that $\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot e^{-x^2}\sin^2(x^2)=\sqrt{\pi}\left(\sqrt{\phi}-1\right)?$

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 $(1)?$

• I have another result – Dr. Sonnhard Graubner Apr 1 '17 at 7:10
• I think the OP result is correct. – Mathxx Apr 1 '17 at 7:12
• You have different form but same result @Dr.Sonnhard – gymbvghjkgkjkhgfkl Apr 1 '17 at 7:12
• The integral in (2) diverges, does it not? (I suppose you want $|u|$, not $u$). – uniquesolution Apr 1 '17 at 7:13
• in (2), the interval being integrated should be $(0,\infty)$ instead of the whole $\Bbb R$ – Nick Apr 1 '17 at 7:18

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.