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

Question

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

Answer 1

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.

Answer 2

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, $$ \begin{align} \int_{-\infty}^\infty\frac{e^{-a^2x^2}}{1+x^2}\,\mathrm{d}x &=f\!\left(a^2\right)\\ &=\pi e^{a^2}\erfc(a)\tag5 \end{align} $$

Answer 3

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

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