I have a feel this will be a duplicate question. I have had a look around and couldn't find it, so please advise if so.

Here I wish to address the definite integral:

\begin{equation} I = \int_{0}^{\infty} \frac{e^{-x^2}}{x^2 + 1}\:dx \end{equation}

I have solved it using Feynman's Trick, however I feel it's limited and am hoping to find other methods to solve. Without using Residues, what are some other approaches to this integral?

My method:

\begin{equation} I(t) = \int_{0}^{\infty} \frac{e^{-tx^2}}{x^2 + 1}\:dx \end{equation}

Here $I = I(1)$ and $I(0) = \frac{\pi}{2}$. Take the derivative under the curve with respect to '$t$' to achieve:

\begin{align} I'(t) &= \int_{0}^{\infty} \frac{-x^2e^{-tx^2}}{x^2 + 1}\:dx = -\int_{0}^{\infty} \frac{x^2e^{-tx^2}}{x^2 + 1}\:dx \\ &= -\left[\int_{0}^{\infty} \frac{\left(x^2 + 1 - 1\right)e^{-tx^2}}{x^2 + 1}\:dx \right] \\ &= -\int_{0}^{\infty} e^{-tx^2}\:dx + \int_{0}^{\infty} \frac{e^{-tx^2}}{x^2 + 1}\:dx \\ &= -\frac{\sqrt{\pi}}{2}\frac{1}{\sqrt{t}} + I(t) \end{align}

And so we arrive at the differential equation:

\begin{equation} I'(t) - I(t) = -\frac{\sqrt{\pi}}{2}\frac{1}{\sqrt{t}} \end{equation}

Which yields the solution:

\begin{equation} I(t) = \frac{\pi}{2}e^t\operatorname{erfc}\left(t\right) \end{equation}


\begin{equation} I = I(1) \int_{0}^{\infty} \frac{e^{-x^2}}{x^2 + 1}\:dx = \frac{\pi}{2}e\operatorname{erfc}(1) \end{equation}


Using the exact method I've employed, you can extend the above integral into a more genealised form:

\begin{equation} I = \int_{0}^{\infty} \frac{e^{-kx^2}}{x^2 + 1}\:dx = \frac{\pi}{2}e^k\operatorname{erfc}(\sqrt{k}) \end{equation}

Addendum 2: Whilst we are genealising: \begin{equation} I = \int_{0}^{\infty} \frac{e^{-kx^2}}{ax^2 + b}\:dx = \frac{\pi}{2b}e^\Phi\operatorname{erfc}(\sqrt{\Phi}) \end{equation}

Where $\Phi = \frac{kb}{a}$ and $a,b,k \in \mathbb{R}^{+}$

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    $\begingroup$ something is wrong. $I<\int_0^\infty e^{-x^2}=\sqrt\pi/2$. Your answer is greater than this value $\endgroup$ – Andrei Dec 16 '18 at 2:54
  • $\begingroup$ @Andrei - You are indeed correct, I mistyped. It should be $\operatorname{efrc}$ not $\operatorname{erf}$. Thank you for the pickup. $\endgroup$ – user150203 Dec 16 '18 at 2:55

You can use Plancherel's theorem. Note that $$ 2I = \int_{-\infty}^{\infty} \frac{e^{-x^2}}{x^2 + 1}dx. $$Let $f(x) = e^{-x^2}$ and $g(x) = \frac{1}{1+x^2}$. Then we have $$ \widehat{f}(\xi) = \sqrt{\pi}e^{-\pi^2\xi^2}, $$ and $$ \widehat{g}(\xi) = \pi e^{-2\pi|\xi|}. $$ By Plancherel's theorem, we have $$\begin{eqnarray} \int_{-\infty}^{\infty} f(x)g(x)dx&=&\int_{-\infty}^{\infty} \widehat{f}(\xi)\widehat{g}(\xi)d\xi\\&=&\pi^{\frac{3}{2}}\int_{-\infty}^{\infty}e^{-\pi^2\xi^2-2\pi|\xi|}d\xi\\ &=&2\pi^{\frac{3}{2}}\int_{0}^{\infty}e^{-\pi^2\xi^2-2\pi\xi}d\xi\\ &=&2\pi^{\frac{3}{2}}e\int_{\frac{1}{\pi}}^{\infty}e^{-\pi^2\xi^2}d\xi\\ &=&2\pi^{\frac{1}{2}}e\int_{1}^{\infty}e^{-\xi^2}d\xi = \pi e \operatorname{erfc}(1). \end{eqnarray}$$ This gives $I = \frac{\pi}{2}e \operatorname{erfc}(1).$

  • $\begingroup$ Thanks @Song! I was not aware of Plancherel's theorem :-) $\endgroup$ – user150203 Dec 16 '18 at 3:17
  • $\begingroup$ With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution). $\endgroup$ – user150203 Dec 20 '18 at 0:38
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    $\begingroup$ Your concern is legitimate. Perhaps the quickest way to evaluate $\widehat{f}$ and $\widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $\int_{-\infty}^\infty e^{zx}e^{-x^2}dx,\;z\in\mathbb{C}$ easily using power series expansion. About $\widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $\int_{-\infty}^\infty\widehat{g}(\xi)e^{2 \pi i\xi x}d\xi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-\infty,0)$ and $(0,\infty)$. $\endgroup$ – Myeonghyeon Song Dec 20 '18 at 2:20

Here is a method that employs the old trick of converting the integral into a double integral.

Observe that $$\frac{1}{1 + x^2} = \int_0^\infty e^{-u(1 + x^2)} \, du.$$ So your integral can be rewritten as $$I = \int_0^\infty e^{-x^2} \int_0^\infty e^{-u(1 + x^2)} \, du \, dx.$$ or $$I = \int_0^\infty e^{-u} \int_0^\infty e^{-(1 + u)x^2} \, dx \, du,$$ on changing the order of integration.

Enforcing a substitution of $x \mapsto x/\sqrt{1 + u}$ gives $$I = \int_0^\infty \frac{e^{-u}}{\sqrt{1 + u}} \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{-u}}{\sqrt{1 + u}} \, du.$$

Next exforcing a substitution of $u \mapsto u^2 - 1$ gives $$I = \sqrt{\pi} e \int_1^\infty e^{-u^2} \, du = \sqrt{\pi} e \cdot \frac{\sqrt{\pi}}{2} \text{erfc} (1) = \frac{\pi e}{2} \text{erfc} (1),$$ as expected.

  • $\begingroup$ Love it. Still developing my skills with respect to integral ‘expansions’. $\endgroup$ – user150203 Dec 16 '18 at 10:00
  • $\begingroup$ Is another one of those good techniques to have handy in the integration toolbox. $\endgroup$ – omegadot Dec 16 '18 at 11:11
  • $\begingroup$ Absolutely. Great technique. $\endgroup$ – user150203 Dec 16 '18 at 11:12
  • $\begingroup$ (+1) Nice approach, but I believe the result should actually be $\frac{\pi e}{2}\color{red}{\mathrm{erfc}}(1)=\frac{\pi e}{2}(1-\mathrm{erf}(1))$ (it's just the last line). $\endgroup$ – projectilemotion Jan 29 at 13:46
  • $\begingroup$ @projectilemotion - Indeed and thanks. $\endgroup$ – omegadot Jan 30 at 7:07

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