# Calculate $\int_{-\infty}^\infty \frac{e^{-x^2}}{x^2+a^2}\ dx$. [duplicate]

Let $$F(a)=\int_{-\infty}^\infty \frac{e^{-x^2}}{x^2+a^2}\ dx, \quad a>0.$$ Is it possible to relate $$F(a)$$ to some known (special) functions?

Parameterize this integral by adding a second parameter, $$t$$: $$I(t):= \int_{-\infty}^\infty \frac{e^{-(x^2+a^2)t}}{x^2+a^2}dx$$ Differentiating with respect to $$t$$, we have $$I'(t)=-\int_{-\infty}^\infty e^{-(x^2+a^2)t}dx=-\sqrt{\frac{\pi}{t}} e^{-a^2 t}$$ This shows us that \begin{align} I(t) &=I(0)-\int_0^t \sqrt{\frac{\pi}{x}}e^{-a^2 x}dx\\ &=\frac{\pi}{a}-2\int_0^{\sqrt{t}} \sqrt{\pi}e^{-a^2 x^2}dx\\ &=\frac{\pi}{a}-\frac{\pi \text{erf}(a\sqrt{t})}{a}\\ &=\frac{\pi \text{erfc}(a\sqrt{t})}{a}\\ \end{align} Which gives us the desired value of your integral: $$\int_{-\infty}^\infty \frac{e^{-x^2}}{x^2+a^2}dx=\frac{\pi e^{a^2}\text{erfc}(a)}{a}$$ Delicious!
As an alternative we can use the following property of the Fourier transform:$$\int_{-\infty}^{+\infty}f(t)g(t)\,dt = \int_{-\infty}^{+\infty}(\mathscr{F} f)(s)(\mathscr{F}^{-1}g)(s)\,ds$$ Simplifying we have: $$\int_{-\infty}^\infty e^{-x^2}\frac{1}{x^2+a^2 } dx=2\int_{0}^\infty\left(\frac{e^{-x^2/4}}{\sqrt 2}\right)\left(\frac{\sqrt{\pi/2} e^{-ax} }{a}\right)dx=$$$$=\frac{\sqrt{\pi}}{a}\int_0^\infty e^{-(x^2/4+ax) } dx=\frac{\sqrt{\pi}}{a}\sqrt{\pi} e^{a^2} \text{erf}\left(a+\frac{x} {2} \right)\bigg|_0^\infty=$$$$=\frac{\pi e^{a^2} } {a}\left(1-\text{erf}(a)\right)=\color{blue}{\frac{ \pi e^{a^2} } {a} \text {erfc}(a)}$$
• What does $G^{-1}(s)$ mean here? – eyeballfrog Dec 18 '18 at 0:47
• The inverse fourier transform of $g(t)$, I don't know if that is the proper notation. I have learned about it from these comments: math.stackexchange.com/a/2891545/515527 – Zacky Dec 18 '18 at 0:49