Evaluatig: $\int_{0}^{\infty}{e^{ax^2}\cos(bx)dx}$

Solution 1:

Assume that $a\lt0$ (otherwise, the integral diverges).

Real Variables

Letting $c=\frac{b}{\sqrt{-a}}$ and $u=\sqrt{-a}\,x$, we get $$ \int_0^\infty e^{ax^2}\cos(bx)\,\mathrm{d}x =\frac1{\sqrt{-a}}\int_0^\infty e^{-u^2}\cos\left(cu\right)\,\mathrm{d}u\tag{1} $$ Now take the derivative with respect to $c$ (which is fine since the integral converges absolutely): $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}c}\int_0^\infty e^{-u^2}\cos\left(cu\right)\,\mathrm{d}u &=-\int_0^\infty e^{-u^2}u\sin\left(cu\right)\,\mathrm{d}u\\ &=\frac12\int_0^\infty\sin\left(cu\right)\,\mathrm{d}e^{-u^2}\\ &=\left.\frac12\sin\left(cu\right)e^{-u^2}\right]_0^\infty-\frac c2\int_0^\infty e^{-u^2}\cos(cu)\,\mathrm{d}u\\ &=-\frac c2\int_0^\infty e^{-u^2}\cos(cu)\,\mathrm{d}u\tag{2} \end{align} $$ That is, $I'(c)=-\frac c2I(c)$. Therefore, $I(c)=\frac{\sqrt\pi}2e^{-\frac{c^2}4}$. Thus, $$ \int_0^\infty e^{ax^2}\cos(bx)\,\mathrm{d}x =\frac{\sqrt\pi}{2\sqrt{-a}}e^{\frac{b^2}{4a}}\tag{3} $$

Contour Integration $$ \begin{align} \int_0^\infty e^{ax^2}\cos(bx)\,\mathrm{d}x &=\frac12\int_{-\infty}^\infty e^{ax^2}\cos(bx)\,\mathrm{d}x\\ &=\frac12\mathrm{Re}\left(e^{\frac{b^2}{4a}}\int_{-\infty}^\infty e^{a\left(x+i\frac{b}{2a}\right)^2}\,\mathrm{d}x\right)\tag{4} \end{align} $$ Use the contour $$ \gamma+R=[-R,R]\cup\color{#C00000}{[R,R+i\frac{b}{2a}]}\cup[R+i\frac{b}{2a},-R+i\frac{b}{2a}]\cup\color{#C00000}{[-R+i\frac{b}{2a},-R]} $$ and Cauchy's Integral Theorem to compute $$ \int_{\gamma_R}e^{az^2}\,\mathrm{d}z=0\tag{5} $$ The integral over the vertical segments (in red) vanishes as $R\to\infty$ and the integral over the horizontal segments (in black) tends to $$ \int_{-\infty}^\infty e^{ax^2}\,\mathrm{d}x-\int_{-\infty}^\infty e^{a\left(x+i\frac{b}{2a}\right)^2}\,\mathrm{d}x\tag{6} $$ Therefore, $(5)$ and $(6)$ imply $$ \begin{align} \int_{-\infty}^\infty e^{a\left(x+i\frac{b}{2a}\right)^2}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{ax^2}\,\mathrm{d}x\\ &=\frac{\sqrt\pi}{\sqrt{-a}}\tag{7} \end{align} $$ Combining $(4)$ and $(7)$ yields $$ \int_0^\infty e^{ax^2}\cos(bx)\,\mathrm{d}x =\frac{\sqrt\pi}{2\sqrt{-a}}e^{\frac{b^2}{4a}}\tag{8} $$

Solution 2:

$$\int_{0}^{\infty}{e^{ax^2}\cos(bx)dx} = \frac{1}{2}\int_{-\infty}^{\infty}{e^{ax^2}\cos(bx)dx}$$ Then you can use $$\cos(bx)=\Re(e^{ibx})$$ To obtain: $$\frac{1}{2}\int_{-\infty}^{\infty}{e^{ax^2}\cos(bx)dx}=\frac{1}{2}\Re\int_{-\infty}^{\infty}{e^{ax^2+ibx}dx}$$ From here you can solve this like a regular Gaussian integral, completing the square and moving the integration variable.

Note: Saying $b>0$ doesn't change anything, as $\cos(bx)$ is an even function.
Also, the integral diverges (obviously, as the function doesn't tend to zero) for $a>0$.