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This question already has an answer here:

I would like to evaluate the following integral $(a>0)$

$$\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx.$$

I've have tried integration by parts, putting $e^{-ax^2}=u$, but I come across with this integral

$$\int_{-\infty}^{\infty}\exp\left(-a x^2\right)\arctan\left(\frac{x}{b}\right),$$

and I don't know how to do it. Could you help me?

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marked as duplicate by Carl Mummert, Adrian Keister, GNUSupporter 8964民主女神 地下教會, Community Oct 3 '18 at 22:42

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    $\begingroup$ In which course did you encounter that integral? Do you know complex analysis? $\endgroup$ – mlainz Oct 3 '18 at 11:48
  • $\begingroup$ After the integration by parts, you have the integral of an odd function over a symmetric interval (though I believe your integration by parts is incorrect).. $\endgroup$ – Mattos Oct 3 '18 at 11:50
  • $\begingroup$ @mlainz This came from my graduate research on Graphene. $\endgroup$ – user1256 Oct 3 '18 at 11:51
  • $\begingroup$ It is $$\frac{\pi e^{a^3} \text{erfc}\left(a^{3/2}\right)}{a}$$ if $$a>0$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 3 '18 at 11:52
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    $\begingroup$ Alternatively it seems that we could compute this by differentiating under the integral sign. $\endgroup$ – xbh Oct 3 '18 at 11:52
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Hint

By using the Schwinger parametrization

$$\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx=\int_{-\infty}^{\infty}dx\, e^{-ax^2}\int_{0}^{\infty}dt\,e^{-t(x^2+b^2)}.$$

The integral over $x-$ variable is a Gaussian integral. To evaluate the integral over $t-$ variable look at the Error function.

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  • $\begingroup$ I really liked this answer. Thank you. $\endgroup$ – user1256 Oct 3 '18 at 22:45
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Call your integral $f(a,\,b)$ so $f(0,\,b)=\frac{\pi}{b}$ and $\partial_a f+Pf=Q$ with $P=-b^2,\,Q=-\sqrt{\pi}a^{-1/2}$. Defining $R:=\exp\int P da=\exp (-ab^2)$,$$f=R^{-1}\int RQ da=-\sqrt{\pi}\exp (ab^2)\int a^{-1/2}\exp (-ab^2)da.$$Getting the integration constant right and substituting $a=b^2c^2$, $$f=\frac{\pi}{b}-\frac{2\sqrt{\pi}}{b}\exp (ab^2)\int_0^{b\sqrt{a}} \exp (-c^2)dc.$$As Dr. Sonnhard Graubner noted, we can rewrite this in terms of the error function.

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  • $\begingroup$ This is very nice. Thank you. $\endgroup$ – user1256 Oct 3 '18 at 22:46
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Differentiating under the integral sign:

Clearly the integral is just $$ \DeclareMathOperator{\diff}{\,d\!} 2 \int_0^{+\infty} \frac {\exp(-ax^2)}{x^2+b^2} \diff x. $$ Let the integral be $I(a)$ and we regard $b$ as a constant. We want to differentiate w.r.t. $a$. For each $c > 0$, on $J_c=[c, +\infty)$, the integral converges for $a \in J_c$, and $$\newcommand{\Abs}[1]{{\left|#1\right|}} \Abs{\partial_a \frac {\exp(-ax^2)}{x^2+b^2}} \leqslant \exp(-cx^2), $$ where $\int_0^{+\infty} \exp(-cx^2) \diff x$ converges, so by Weierstrass M-test, $$ \int_0^{+\infty} \frac {\exp(-ax^2)(-x^2)}{x^2+b^2}\diff x $$ converges uniformly for $a\in J_c$. Therefore $I(a)$ could be differentiated under the integral symbol. Now \begin{align*} I'(a) &= \int_0^{+\infty} \frac {\exp(-ax^2) (-x^2)}{x^2+b^2}\diff x\\ &= \int_0^{+\infty} \exp(-ax^2) \left( \frac {b^2}{x^2+b^2}-1 \right)\diff x\\ &= b^2 I(a) - \int_0^{+\infty} \exp(-ax^2)\diff x\\ &= b^2 I(a) - \frac {\sqrt \pi} {2\sqrt a}. \end{align*} Now solve this differential equations under the initial value condition $$ I(0) = \int_0^{+\infty}\frac {\diff x}{x^2+b^2} = \frac \pi{2b}. $$ By the formula for 1st-order ODE, we have $$ I(a) = \frac{\exp(b^2a)}b \left(-\int_0^{b\sqrt a} \exp(-u^2)\diff u + \frac \pi {2b}\right), $$ and the original is just $2I(a)$.

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  • $\begingroup$ Seems spent too many time on typing… $\endgroup$ – xbh Oct 3 '18 at 12:28
  • $\begingroup$ Nice. Thank you for your answer. $\endgroup$ – user1256 Oct 3 '18 at 22:47

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