For $\alpha>0$, evaluate $\int^{+\infty}_0xe^{-x}\cos x\cos(x^2/\alpha)\,dx$ For $\alpha>0$, prove\begin{align}\displaystyle\int_0^{\infty}xe^{-x}\cos(x)\cos\left(\dfrac{x^2}{\alpha}\right){\rm d}x=\dfrac{\alpha\sqrt{2\pi\alpha}}{8}e^{-\alpha/2} \end{align}
My attempt: Let $$I(b)=\displaystyle\int_0^{\infty}e^{-x}\sin(bx)\cos\left(\dfrac{x^2}{\alpha}\right){\rm d}x$$
Hence$$I'(1)=\displaystyle\int_0^{\infty}xe^{-x}\cos(x)\cos\left(\dfrac{x^2}{\alpha}\right){\rm d}x  $$
But \begin{align}I(b) & =\displaystyle\int_0^{\infty}e^{-x}\sin(bx)\cos\left(\dfrac{x^2}{\alpha}\right){\rm d}x\\&=\dfrac{1}{2}\displaystyle\int_0^{\infty}e^{-x}\left(\sin(bx-x^2/\alpha)+\sin(bx+x^2\alpha)\right){\rm d}x\\&=I_1+I_2\end{align}
But I have difficulty using contour integral to calculate $I_1$ or $I_2$ .
how to solve it using contour inregral?
Is there a more efficient method to solve this problem? I have a thought that Fourier Transform is a possible method.
 A: I will write
$$ J_{\pm} = \int_{0}^{\infty} xe^{-x}\cos\left(x\pm\frac{x^2}{\alpha}\right)\,\mathrm{d}x $$
so that your integral takes the form $\frac{1}{2}(J_{+} + J_{-})$. Then
\begin{align*}
J_{\pm}
&= \operatorname{Re}\left[ \int_{0}^{\infty} x\exp\left( -x + ix \pm \frac{ix^2}{\alpha}\right) \,\mathrm{d}x. \right]
\end{align*}
Now write $\mathbb{H}_{\text{right}} = \{ z \in \mathbb{C} : \operatorname{Re}(z) > 0 \}$. Then for each $a \in \mathbb{H}_{\text{right}}$, the map $z \mapsto \int_{0}^{\infty} x e^{-ax-zx^2} \, \mathrm{d}x$ is analytic on $\mathbb{H}_{\text{right}}$ and continuous on $\overline{\mathbb{H}_{\text{right}}}$. Moreover, if $a, z \in (0, \infty)$, then with $b = a^2/4z$,
\begin{align*}
\int_{0}^{\infty} x e^{-ax-zx^2} \, \mathrm{d}x
&= \int_{0}^{\infty} x \exp\bigg( -b \left(\frac{2z x}{a}+1\right)^2 + b \bigg) \, \mathrm{d}x \\
&= \frac{be^{b}}{z} \int_{1}^{\infty} (u-1) e^{-bu^2} \, \mathrm{d}u \\
&= \frac{be^{b}}{z} \left( \int_{1}^{\infty} u e^{-bu^2} \, \mathrm{d}u - \int_{0}^{\infty} e^{-bu^2} \, \mathrm{d}u + \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u \right) \\
&= \frac{1}{2z} - \frac{\sqrt{\pi} a e^{b}}{4z^{3/2}} + \frac{be^{b}}{z} \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u.
\end{align*}
Then by the principle of analytic continuation, this holds for all $a \in \mathbb{H}_{\text{right}}$ and $z \in \overline{\mathbb{H}_{\text{right}}}$. Then plugging $a = 1-i$ and $z = z_{\pm} = \pm i/\alpha$, we get $b = b_{\pm} = \mp \alpha /2 \in \mathbb{R}$. 
\begin{align*}
J_{\pm}
&= \operatorname{Re}\bigg[ \frac{1}{2z} - \frac{\sqrt{\pi} a e^{b}}{4z^{3/2}} + \frac{be^{b}}{z} \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u \bigg] \\
&= -\frac{\sqrt{\pi} \, e^{b}}{4} \operatorname{Re}\bigg[ \frac{a}{z^{3/2}} \bigg],
\end{align*}
By noting that $a_{+}/z_{+}^{3/2} = -\sqrt{2}\,\alpha^{3/2}$ and $a_{-}/z_{-}^{3/2} = i\sqrt{2}\,\alpha^{3/2}$, we get
$$ J_{+} = \frac{\sqrt{2\pi}}{4} \alpha^{3/2} e^{-\alpha/2}, \qquad J_{-} = 0. $$
This complete the proof.
A: Let $$I(\beta) =\int_{0}^{\infty} x e^{-x} \cos (x) \cos( \beta x^{2}) \, \mathrm dx \, , \quad \beta>0. $$
Let's take the Laplace transform of $I (\beta)$ and then switch the order of integration (which is permissible since the iterated integral converges absolutely).
$$ \begin{align}\mathcal{L} \{I(\beta)\} (s) &= \int_{0}^{\infty} \left(\int_{0}^{\infty} x e^{-x} \cos(x) \cos(\beta x^{2}) \, \mathrm dx \right)e^{-s \beta } \,  \mathrm d \beta \\ &= \int_{0}^{\infty} x e^{-x} \cos(x) \int_{0}^{\infty}\cos(\beta x^{2}) e^{- s \beta } \, \mathrm d \beta \, \mathrm dx \\ &= s \int_{0}^{\infty} \frac{x e^{-x}\cos (x)}{x^{4}+s^{2}} \, \mathrm dx \tag{1} \end{align} $$
To evaluate $(1)$, we can integrate the complex function $$f(z) = \frac{z e^{-z}e^{iz}}{z^{4}+s^{2}} $$ around a wedge-shaped contour that makes an angle of $\frac{\pi}{2}$ with the positive real axis (i.e., a closed quarter-circle in the first quadrant of the complex plane).
In the first quadrant of the complex plane, $\vert e^{-z} \vert \le 1$ and $\vert e^{iz} \vert \le 1$.  So the integral along the big arc clearly vanishes as the radius of the arc goes to $\infty$. 
Integrating around the contour, we get $$ \begin{align} \int_{0}^{\infty} \frac{x e^{-x} e^{ix}}{x^{4}+s^{2}} \, \mathrm dx + \int_{\infty}^{0} \frac{(it)e^{-it} e^{-t}}{t^{4}+s^{2}} \, i \, \mathrm dt &= 2 \pi i \operatorname{Res} \left[f(z), e^{i \pi/4} \sqrt{s}  \right] \\ &= 2 \pi i \lim_{z \to e^{i \pi/4} \sqrt{s}}\frac{ze^{-z}e^{iz}}{4z^{3}} \\ &= 2 \pi i \, \frac{e^{i \pi/4}\sqrt{s} \, e^{-\sqrt{2s}} }{4e^{3 \pi i/4}s^{3/2}} \\ &= \pi \, \frac{e^{-\sqrt{2s}}}{2s}. \end{align}$$
And if we equate the real parts on both sides of the equation, we get $$2 \int_{0}^{\infty} \frac{xe^{-x} \cos(x) }{x^{4}+s^{2} } \, \mathrm dx = \pi \,  \frac{e^{-\sqrt{2s}}}{2s}.$$
Therefore, $$\mathcal{L} \{I(\beta)\} (s) = \frac{\pi e^{-\sqrt{2s}}}{4}. $$
Due to the uniqueness of the inverse Laplace transform , we only need to show that  $$\mathcal{L} \left\{\frac{\sqrt{2 \pi}}{8} \beta^{-3/2} e^{-1/(2 \beta)} \right\} (s) = \frac{\pi e^{-\sqrt{2s}}}{4}$$ in order to prove that $$I(\beta) =  \frac{\sqrt{2 \pi}}{8} \beta^{-3/2} e^{-1/(2 \beta)}.$$
From the answers here, we know that for $a,b>0$, $$\int_{0}^{\infty} \frac{\exp \left(-ax^{2}-b/x^{2}\right)}{x^{2}} \, \mathrm dx =  - \frac{1}{2} \sqrt{\frac{\pi}{a}} \, \frac{\partial }{\partial  b} e^{-2\sqrt{ab}} = \frac{1}{2} \sqrt{\frac{\pi}{b}} \, e^{-2\sqrt{ab}}.$$
(Differentiation under the integral sign is permissible since for any positive $c$ less than $b$, $\frac{\exp \left(-ax^{2}-b/x^{2}\right)}{x^{2}}$ is dominated by the integrable function $ \frac{\exp \left(-ax^{2}-c/x^{2}\right)}{x^{2}}$.)
Therefore, $$\begin{align} \mathcal{L} \left\{\frac{\sqrt{2 \pi}}{8} \beta^{-3/2} e^{-1/(2 \beta)} \right\} (s) &=  \frac{\sqrt{2 \pi}}{8} \int_{0}^{\infty}\beta^{-3/2} e^{-1/(2 \beta)} e^{-s \beta } \, \mathrm d \beta \\ &= \frac{\sqrt{2 \pi}}{4} \int_{0}^{\infty} \frac{\exp \left(-s u^{2} - (1/2)/u^{2} \right)}{u^{2}} \mathrm du \\ &= \frac{\sqrt{2 \pi}}{4}  \frac{\sqrt{2 \pi}}{2} \, e^{-\sqrt{2s}}  \\ &= \frac{ \pi \, e^{-\sqrt{2s}}}{4}. \end{align}$$
