# Integral of the form $\int_{-\infty}^{\infty} \frac{e^{-(ax)^2}}{1 + x^2}dx$

I was reading a paper on nuclear physics when I came across the following definite integral:

$$\int_{-\infty}^{\infty}\frac{\zeta}{2\sqrt\pi} \frac{e^{-\frac{\zeta^2}{4} y^2}}{1 + y^2}\mathrm dy$$

The paper gives the expression of the above integral as:

$$\int_{-\infty}^{\infty}\frac{\zeta}{2\sqrt\pi} \frac{e^{-\frac{\zeta^2}{4} y^2}}{1 + y^2}\mathrm dy = \frac{\zeta \sqrt\pi}{2} e^{\frac{\zeta^2}{4}}\left(1-\operatorname{erf}\left (\frac{\zeta}{2}\right )\right)$$

Basically, I have no clue where this result comes from. I have tried the substitution $$u = \tan^{-1}y$$ so that $$\mathrm du = \frac{1}{1 + y^2}\mathrm dy$$, but I get the following expression:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\zeta}{2\sqrt\pi} e^{-\frac{\zeta^2}{4} \tan^2(u)}\mathrm du$$

Which I do not know how to evaluate. Any help on the above integral would be greatly appreciated. Even a hint as to how to proceed further is welcomed. Thank you so much in advance!

PS: This is my first question so I hope the formatting/question wording is not too confusing.

Best,

Nathan

• Are you familiar with derivatives under the integral sign? – Nyssa Sep 3 at 14:01
• Hint: Let $I(\zeta)=\int_{\mathbb{R}}{e^{-\zeta^2y^2/4}\frac{dy}{1+y^2}}$, then $\frac{d}{d\zeta}(e^{-\zeta^2/4}I(\zeta))= ?$ – Mindlack Sep 3 at 14:04
• @Zacky Yes I am though vaguely, I am from an Engineering background so my Maths knowledge is rather limited. – NRavoisin Sep 3 at 14:05
• You may consider the Fourier transforms of $e^{-ax^2}$ (still is a Gaussian function) and $\frac{1}{1+x^2}$ (you get a Laplace distribution), complete the square and profit. – Jack D'Aurizio Sep 3 at 17:48

Consider the following integral: $$I(a)=\int_{-\infty}^\infty \frac{e^{-a^2(1+x^2)}}{1+x^2}dx$$ Note that initially the constant $$e^{-a^2}$$ wasn't there, but bringing it helps to simplify the denominator when we take a derivative with respect to $$a$$. Afterwards we just mutiply by $$e^{a^2}$$ and everything is unchanged, but let's take a derivative: $$I'(a)=-2a\int_{-\infty}^\infty e^{-a^2(1+x^2)}dx=-2\sqrt \pi e^{-a^2}$$ Now notice that $$I(\infty)=0$$ and we're after $$I\left(\frac{\zeta }{2}\right)$$. $$I\left(\frac{\zeta }{2}\right)=-\left(I(\infty)-I\left(\frac{\zeta}{2}\right)\right)=2\sqrt \pi \int_{\frac{\zeta}{2}}^\infty e^{-a^2}da=\pi\operatorname{erfc}\left(\frac{\zeta }{2}\right)$$ Finally we just need to multiply by $$\frac{\zeta }{2\sqrt \pi}e^{\zeta^2/4}$$ and the result follows.

Define $$f(a)=\int_{-\infty}^\infty\frac{e^{-ax^2}}{1+x^2}\,\mathrm{d}x\tag1$$ then \begin{align} f(a)-f'(a) &=\int_{-\infty}^\infty\frac{e^{-ax^2}}{1+x^2}\,\mathrm{d}x-\frac{\mathrm{d}}{\mathrm{d}a}\int_{-\infty}^\infty\frac{e^{-ax^2}}{1+x^2}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty e^{-ax^2}\,\mathrm{d}x\\ &=\sqrt{\frac\pi{a}}\tag2 \end{align} We can solve $$(2)$$ using an integrating factor. Note that \begin{align} \left(e^{-a}f(a)\right)' &=-e^{-a}f(a)+e^{-a}f'(a)\\[3pt] &=-e^{-a}(f(a)-f'(a))\\ &=-e^{-a}\sqrt{\frac\pi{a}}\tag3 \end{align} Therefore, using the complementary error function, \newcommand{\erfc}{\operatorname{erfc}} \begin{align} f(a) &=e^a\int_a^\infty e^{-t}\sqrt{\frac\pi{t}}\,\mathrm{d}t\\ &=2\sqrt\pi e^a\int_{\sqrt{a}}^\infty e^{-t^2}\,\mathrm{d}t\\ &=\pi e^a\erfc\left(\sqrt{a}\right)\tag4 \end{align} Thus, \begin{align} \int_{-\infty}^\infty\frac{e^{-a^2x^2}}{1+x^2}\,\mathrm{d}x &=f\!\left(a^2\right)\\ &=\pi e^{a^2}\erfc(a)\tag5 \end{align}

Starting with

$$\Re\int_0^\infty e^{-(1+ix)y}\ dy=\Re\frac1{1+ix}=\frac1{1+x^2}\tag{1}$$

$$\int_0^\infty e^{-(ax^2+bx+c)}\ dx=\frac12\sqrt{\frac{\pi}{a}}\ e^{\frac{b^2}{4a}-c}\ \text{erfc}\left(\frac{b}{2\sqrt{a}}\right)\tag{2}$$

Multiply both sides of (1) by $$e^{-a^2x^2}$$ then integrate from $$x=0$$ to $$\infty$$ we have

\begin{align} \int_0^\infty\frac{e^{-a^2x^2}}{1+x^2}\ dx&=\int_0^\infty e^{-y}\left(\Re \int_0^\infty e^{-(a^2x^2+iyx)}\ dx\right)\ dy\\ &\overset{\text{use (2)}}{=}\int_0^\infty e^{-y}\left(\frac{\sqrt{\pi}}{2a}e^{-\frac{y^2}{4a^2}}\right)\ dy\\ &=\frac{\sqrt{\pi}}{2a}\int_0^\infty e^{-(\frac{y^2}{4a^2}+y)}\ dy\\ &\overset{\text{use (2)}}{=}\frac{\sqrt{\pi}}{2a}\left(a\sqrt{\pi}e^{a^2}\text{erfc}(a)\right)\\ &=\frac{\pi}{2}\ e^{a^2}\text{erfc}(a) \end{align}

and since the integrand is even function, then

$$\int_{-\infty}^\infty\frac{e^{-a^2x^2}}{1+x^2}\ dx=2\int_0^\infty\frac{e^{-a^2x^2}}{1+x^2}\ dx=\pi\ e^{a^2}\text{erfc}(a)$$