How to compute $\int_{0}^{\infty}e^{-ax^2}\cos(bx^2)dx$?

Question

How to compute $\int_{0}^{\infty}e^{-ax^2}\cos(bx^2)dx$? I'm seeing gaussian integral here. I tried different methods, but nothing helped

Answer 1

Your integral can be re-written as $$\frac{1}{2}\text{Re} \int_{-\infty}^{\infty} e^{-(a-ib)x^2} \, dx.$$ But this is just a Gaussian integral, albeit with a complex exponential, whose result is famous, and in this case is given by $$\frac{1}{2} \cos(\phi/2)\sqrt{\frac{\pi}{(a^2+b^2)^{1/2}}},$$ where $\phi = \text{arg}\,z$, for $z = a - ib$.

Answer 2

I assume that $a > 0$.

One approach is to use a contour integral. First, by symmetry note that $$ I = \int_{0}^{\infty}e^{-ax^2}\cos(bx^2)dx = \operatorname{Re}\left[ \int_{0}^{\infty}e^{(-a + bi)x^2}\,dx \right]. $$ Let $\omega = a - bi$. We have $$ \int_{0}^{\infty}e^{(-a + bi)x^2}\,dx = \int_{\Bbb R_+} e^{-\omega z^2}\,dz. $$ With the substitution $u = \omega^{1/2} z$, we rewrite this as $$ \omega^{-1/2}\int_{\omega \Bbb R_+} e^{-u^2}\,du. $$ By considering a triangular contour, we find that $\int_{\omega \Bbb R} e^{-u^2}\,du = \int_{\Bbb R} e^{-u^2}\,du$. With that, we find $$ I = \operatorname{Re}\left[ \omega^{-1/2}\int_{0}^{\infty}e^{-x^2}\,dx \right] = \operatorname{Re}(\omega^{-1/2})\frac{\sqrt{\pi}}{2} =\frac{1}{2} \cos(\phi/2)\sqrt{\frac{\pi}{(a^2+b^2)^{1/2}}}, $$ where $\phi = \arg(\omega)$. This confirms the other answer's result.