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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 $\endgroup$ Commented Apr 1, 2017 at 7:10
  • $\begingroup$ I think the OP result is correct. $\endgroup$
    – Mathxx
    Commented Apr 1, 2017 at 7:12
  • $\begingroup$ You have different form but same result @Dr.Sonnhard $\endgroup$ Commented Apr 1, 2017 at 7:12
  • $\begingroup$ The integral in (2) diverges, does it not? (I suppose you want $|u|$, not $u$). $\endgroup$ Commented Apr 1, 2017 at 7:13
  • $\begingroup$ in (2), the interval being integrated should be $(0,\infty)$ instead of the whole $\Bbb R$ $\endgroup$
    – Nick
    Commented Apr 1, 2017 at 7:18

3 Answers 3

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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. $\endgroup$ Commented Apr 1, 2017 at 8:22
  • $\begingroup$ @ClaudeLeibovici You are very welcome. Thank you! $\endgroup$ Commented Apr 1, 2017 at 11:04
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Note $$\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot e^{-x^2}\sin^2(x^2)=2\int_{0}^{+\infty}{\mathrm dx\over x^2}\cdot e^{-x^2}\sin^2(x^2)\overset{x\to x^2}=\int_{0}^{+\infty}{\mathrm dx\over x^{3/2}}\cdot e^{-x}\sin^2(x). $$ Noting $$ \mathcal{L}\{s^{\frac12}\}(x)=\frac{\sqrt{\pi}}{2x^{\frac32}}, \mathcal{L}\{e^{-x}\sin^2(x)\}(s)=\frac2{(s+1)(s^2+2s+5)} $$ one has \begin{eqnarray} &&\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\cdot e^{-x^2}\sin^2(x^2)\\ &=&\int_{0}^{+\infty}{\mathrm dx\over x^{3/2}}\cdot e^{-x}\sin^2(x)\\ &=&\frac{2}{\sqrt\pi}\int_{0}^{+\infty}e^{-x}\sin^2(x)\mathcal{L}\{s^{\frac12}\}(x)\mathrm dx=\frac{2}{\sqrt\pi}\int_{0}^{+\infty}\mathcal{L}\{e^{-x}\sin^2(x)\}(s)s^{\frac12}\mathrm ds\\ &=&\frac{2}{\sqrt\pi}\int_{0}^{+\infty}\frac{2s^{\frac12}}{(s+1)(s^2+2s+5)}\mathrm ds=\overset{s\to s^2}=\frac{8}{\sqrt\pi}\int_{0}^{+\infty}\frac{s^2}{(s^2+1)(s^4+2s^2+5)}\mathrm ds\\ &=&\frac{8}{\sqrt\pi}\int_{0}^{+\infty}\frac{s^2}{(s^2+1)(s^4+2s^2+5)}\mathrm ds=\frac{2}{\sqrt\pi}\int_{0}^{+\infty}\bigg(-\frac1{s^2+1}+\frac{s^2+5}{s^4+2s^2+5}\bigg)\mathrm ds\\ &=&\frac{2}{\sqrt\pi}\bigg(-\frac\pi2+\frac12\sqrt\phi\pi\bigg)=\sqrt\pi(\sqrt\phi-1). \end{eqnarray}

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Making the subsution $u = x^{2}$ and using a trig identity, we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{e^{-x^{2}} \sin^{2}(x^{2})}{x^{2}} \, \mathrm dx &= 2 \int_{0}^{\infty} \frac{e^{-x^{2}} \sin^{2}(x^{2})}{x^{2}} \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{e^{-u}\sin^{2}(u)}{u^{3/2}} \, \mathrm du \\ & = \frac{1}{2} \int_{0}^{\infty} \frac{e^{-u} \left(1-\cos(2u) \right)}{u^{3/2}} \, \mathrm du. \end{align}$$

For $\Re(s) >0$, we have $$ \begin{align} \int_{0}^{\infty} e^{-u} \left(1- \cos(2u) \right) u^{s-1} \, \mathrm du &= \int_{0}^{\infty}e^{-u} u^{s-1} \, \mathrm du - \int_{0}^{\infty}e^{-u} \cos(2u) u^{s-1} \, \mathrm du \\ &\overset{(1)}{=} \Gamma(s) - \frac{\Gamma(s) \cos \left(s \arctan 2 \right)}{5^{s/2}} \\ &= \Gamma(s) \left(1-\frac{\cos \left(s \arctan 2 \right)}{5^{s/2}} \right). \end{align}$$

The Mellin transform $$\int_{0}^{\infty} e^{-u} \left(1- \cos(2u) \right) u^{s-1} \, \mathrm du $$ defines an analytic function for $\Re(s) >-2$.

And the function $$\Gamma(s) \left(1-\frac{\cos \left(s \arctan 2 \right)}{5^{s/2}} \right) $$ can be analytically continued to $\Re(s) >-2$ since it has removable singularities at $s=0$ and $s=-1$.

Therefore, by the identity theorem, $$\int_{0}^{\infty} e^{-u} \left(1- \cos(2u) \right) u^{s-1} \, \mathrm du = \Gamma(s) \left(1-\frac{\cos \left(s \arctan 2 \right)}{5^{s/2}} \right) \, , \quad \Re(s) >-2.$$

So we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{e^{-x^{2}} \sin^{2}(x^{2})}{x^{2}} \, \mathrm dx &= \frac{1}{2} \, \Gamma \left(- \frac{1}{2} \right) \left(1- 5^{1/4} \cos \left(\frac{1}{2} \, \arctan 2 \right)\right) \\ &= -\sqrt{\pi} \left(1-5^{1/4} \sqrt{\frac{1+ \cos(\arctan 2)}{2}} \right) \\ &= -\sqrt{\pi} \left(1-5^{1/4} \sqrt{\frac{1+\frac{1}{\sqrt{5}}}{2}} \right) \\ &= - \sqrt{\pi} \left(1- \sqrt{\frac{\sqrt{5}+1}{2}}\right) \\ &= \sqrt{\pi} \left(\sqrt{\phi} - 1 \right). \end{align}$$


$(1)$ Prove that $\int_{0}^{+\infty} u^{s-1} \cos (a u) \:e^{-b u}\:du=\frac{\Gamma(s)\cos\left(s\arctan \left(\frac{a}{b}\right)\right)}{(a^2+b^2)^{s/2}}$

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