Integral of $\int_0^\infty\sin x/x e^{-x^2/(4T)}dx$ According to Mathematica we have
\begin{equation}
\int_0^\infty\frac{\sin x}{x} e^{-x^2/(4T)}dx=\sqrt{\pi}\int_0^\sqrt{T}e^{-t^2}dt.
\end{equation}
I struggle to see how I can cast the integral in this form. I have tried several substitutions and integration by parts, but cannot get my head around it. Could somebody help with the steps to get there?
Thanks a lot!
 A: The hint about Plancherel's theorem did the trick very nicely. I first define the Fourier transform
\begin{equation}
\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{ikx}dx
\end{equation}
for which Plancherel's identity holds ($f(x)$ and $g(x)$ real),
\begin{equation}
\int_{-\infty}^\infty f(x)g(x)dx=\int_{-\infty}^\infty \hat{f}(k)\hat{g}(k)dk.
\end{equation}
With the identification $f(x) = \frac{\sin x}{x}$, $g(x)=e^{-x^2/(4T)}$ and thus $\hat{f}(k)=\sqrt{\frac{\pi}{2}}\mathrm{rect}(k)$, as well as $\hat{g}(k)=\sqrt{2T}e^{-Tk^2}$ I get
\begin{eqnarray}
\int_0^\infty\frac{\sin x}{x} e^{-x^2/(4T)}dx&=&\frac{1}{2}\int_{-\infty}^\infty\frac{\sin x}{x} e^{-x^2/(4T)}dx\\
&=&\frac{1}{2}\int_{-\infty}^\infty\sqrt{\frac{\pi}{2}}\mathrm{rect}(k) \sqrt{2T}e^{-Tk^2}dk\\
&=&\frac{1}{2}\sqrt{T\pi}\int_{-1}^1 e^{-Tk^2}dk\\
&=&\sqrt{T\pi}\int_{0}^1e^{-Tk^2}dk,
\end{eqnarray}
which is the result I was looking for.
In the beginning and in the end I used that the integrand is symmetric and the rectangular function is defined as
\begin{equation}
\mathrm{rect}(k) = 
\begin{cases}
    1,&  |x|\leq 1\\
    0,          & \text{otherwise}.
\end{cases}
\end{equation}
A: Let
\begin{eqnarray}
I(T)&=&\int_0^\infty\frac{\sin (x)}{x} e^{-x^2/(4T)}dx.
\end{eqnarray}
and then
\begin{eqnarray}
I(\infty)=\frac\pi2, I'(T)&=&\frac{1}{4T^2}\int_0^\infty x\sin (x) e^{-x^2/(4T)}dx.
\end{eqnarray}
Note
\begin{eqnarray} 
J(a)&=&\int_0^\infty \cos (ax) e^{-x^2/(4T)}dx\\
&=&\sqrt{\pi T}e^{-a^2T}
\end{eqnarray}
and hence
\begin{eqnarray}
I'(T)&=&-\frac{1}{4T^2}J'(1)\\
&=&\frac{1}{2T}\sqrt{\pi T}e^{-T}.
\end{eqnarray}
So
$$ I(T)=\int_0^T\frac{1}{2t}\sqrt{\pi t}e^{-t}dt+C=\sqrt{\pi}\int_0^\sqrt{T}e^{-t^2}dt+C$$
Using $I(\infty)=\frac\pi2$, one has $C=0$. Thus
$$ I(T)=\sqrt{\pi}\int_0^\sqrt{T}e^{-t^2}dt. $$
