How to evaluate the integral in the following cosine Fourier transform?

Question

Could anyone prove the following equation: $$\int^{\infty}_0 e^{-a\sqrt{x}}\cos(a\sqrt{x})\cos(bx)\,dx=a\sqrt{\frac{\pi}{8b^3}}e^{-\frac{a^2}{2b}}$$

Answer 1

Here are some first steps:

Let's first do the substitution $\sqrt{x}=u$ so to get an integral without square roots: $dx=2u\,du$ \begin{align}\int^{\infty}_0 e^{-a\sqrt{x}}\cos(a\sqrt{x})\cos(bx)\,dx& = \int^{\infty}_0 e^{-au}\cos(au)\cos(bu^2)2u\,dx\\&=\int^{\infty}_0 e^{-au}\frac{e^{iau}+e^{-iau}}{2}\frac{e^{ibu^2}+e^{-ibu^2}}{2}\cdot 2u\,du\\&=\text{a sum of terms of the form}\\ & = \int_0^\infty u\ e^{c_1u^2+c_2u+c_3}\, du \end{align} where $c_1,c_2,c_3$ are complex numbers.

This integral is equivalent to calculating the first moment of a normal distribution. To evaluate that, we note that $$\frac{d}{du}e^{c_1u^2+c_2u+c_3}=(2uc_1+c_2)e^{c_1u^2+c_2u+c_3},$$ and thus (by matching constants accordingly) rewrite the above as $$\int_0^\infty (2c_1u+c_2)\ e^{c_1u^2+c_2u+c_3}\, du-\int_0^\infty c_2\ e^{c_1u^2+c_2u+c_3}\, du. $$ The first integrand has a simple primitive. The second can be evaluted directly using the formula for the Gaussian integral