Another integral related to Fresnel integrals How would we prove this result by real methods ?
$$\int_0^{\infty } \frac{\sin \left(\pi  x^2\right)}{x+2} \, dx=\frac{1}{4} \left(\pi-2 \pi  C\left(2 \sqrt{2}\right)-2 \pi  S\left(2 \sqrt{2}\right)+2 \text{Si}(4 \pi ) \right)$$
As you can easily see, Fresnel integrals are involved. What are your ideas on it?
 A: Integrating $f(z)=\dfrac{e^{iaz^2}}{z+b}$ along $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$ gives
\begin{align}
\int^\infty_0\frac{\sin(ax^2)}{x+b}\ {\rm d}x
&=\int^\infty_0\frac{b\cos(ay^2)-y\sin(ay^2)}{y^2+b^2}\ {\rm d}y
\end{align}
To compute the first integral, we consider the function
$\displaystyle I(a)=\int^\infty_0\frac{e^{iay^2}}{y^2+b^2}\ {\rm d}y$
 such that $\displaystyle-iI'(a)+b^2I(a)=\frac{\sqrt{\pi}}{2\sqrt{2a}}(1+i)$. Solving this ode while noting the initial value $I(0)=\dfrac{\pi}{2b}$,
\begin{align}
I(a)
&=-e^{-iab^2}\left(\frac{\sqrt{\pi}}{{2\sqrt{2}}}(1-i)\right)\int \frac{\cos(ab^2)+i\sin(ab^2)}{\sqrt{a}}\ {\rm d}a\\
&=-e^{-iab^2}\left(\frac{\sqrt{\pi}}{{2\sqrt{2}}}(1-i)\right)\left[\frac{\sqrt{2\pi}}{b}C\left(b\sqrt{\frac{2a}{\pi}}\right)+i\frac{\sqrt{2\pi}}{b}S\left(b\sqrt{\frac{2a}{\pi}}\right)-\frac{\sqrt{\pi}}{b\sqrt{2}}(1+i)\right]
\end{align}
Taking the real part of $bI(a)$,
\begin{align}
\int^\infty_0\frac{b\cos(ay^2)}{y^2+b^2}\ {\rm d}y
&=\frac{\pi}{2}\bigg{[}C\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)-\cos(ab^2)\right)-S\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)+\cos(ab^2)\right)\\
&\ \ \ \ +\cos(ab^2)\bigg{]}
\end{align}
The second integral readily reduces to sine and cosine integrals.
\begin{align}
\int^\infty_0\frac{y\sin(ay^2)}{y^2+b^2}\ {\rm d}y
&=\frac{1}{2}\int^\infty_{0}\frac{\sin(ay)}{y+b^2}\ {\rm d}y\\
&=\frac{1}{2}\int^\infty_{b^2}\frac{\sin(ay)\cos(ab^2)-\cos(ay)\sin(ab^2)}{y}\ {\rm d}y\\
&=\frac{1}{2}\bigg{[}\operatorname{Si}(ay)\cos(ab^2)-\operatorname{Ci}(ay)\sin(ab^2)\bigg{]}^\infty_{b^2}\\
&=\frac{\pi}{4}\cos(ab^2)-\frac{1}{2}\operatorname{Si}(ab^2)\cos(ab^2)+\frac{1}{2}\operatorname{Ci}(ab^2)\sin(ab^2)
\end{align}
Therefore, we have a generalised result that holds for positive, real $a,b$.
\begin{align}
\color{indigo}{\int^\infty_0\frac{\sin(ax^2)}{x+b}\ {\rm d}x}
&\color{indigo}{=\frac{\pi}{2}\bigg{[}C\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)-\cos(ab^2)\right)-S\left(b\sqrt{\frac{2a}{\pi}}\right)\left(\sin(ab^2)+\cos(ab^2)\right)}\\
&\ \ \ \ \color{indigo}{+\frac{1}{2}\cos(ab^2)\bigg{]}+\frac{1}{2}\left(\operatorname{Si}(ab^2)\cos(ab^2)-\operatorname{Ci}(ab^2)\sin(ab^2)\right)}
\end{align}
Setting $a=\pi$, $b=2$ reproduces the identity stated in the question.
A: Ok, i will give it a shot:
Writing $\int_{0}^{\infty}e^{-t(x+2)}=\frac{1}{x+2}$ and using $\Im(e^{ix})=\sin(x)$ we may reformulate the problem as follows:
$$
I=\Im\left[\int_0^{\infty}dte^{-2 t}\underbrace{\int_0^{\infty}dxe^{i\pi x^2-tx}}_{J(t)}\right]
$$
the inner intgral $J(t)$ is quite straightforward (and also well known because it is just the laplace transform of a gaussian) if one is aware of the definition of the complementary Error function and completes the square.
We get
$$
J(t)=-\frac{(-1)^{3/4}}{2}e^{- a^2 t^2} \text{erfc}\left(i a t\right)
$$
with $a=\frac{(-1)^{3/4}}{2\sqrt{\pi}}$
We therefore left with 
$$
I=\Im\left[-\frac{(-1)^{3/4}}{2}\int_0^{\infty}dte^{-2 t}e^{- a^2 t^2} \text{erfc}\left(i a t\right)\right]
$$ 
To calculate this integral we use $\text{erfc}(z)=1-\text{erf}(z)$ and 4.3.12 in this fantastic paper to obtain:
$$
I=\Im\left[-\frac{(-1)^{3/4}}{4 a}e^\frac{1}{a^2}\left(\sqrt{\pi} \text{erfc}\left(\frac{1}{a}\right)-\frac{1}{i\sqrt{\pi}}\text{Ei}\left(-\frac{1}{a^2}\right)\right) \right]
$$
Here $\text{Ei}(z)$ denotes the exponential integral.
It's now a matter of straightforward but painstaking calculations to get everything in the form you suggested. I'm too lazy for that but instead give a proof of proposition 4.3.12
Proposition
$$
Q(a,b)=\int_0^{\infty}dte^{-b t}e^{- a^2 t^2} \text{erfc}\left(i a t\right)=\frac{1}{2 a i\sqrt{\pi}}e^{\frac{b^2}{4a^2}}\text{Ei}\left(-\frac{b^2}{4a^2}\right)
$$
Proof:
We may use the following respresentation of the error function:
(a proof may be found here)
$$
\text{erf}(z)=\frac{2e^{-z^2}}{\sqrt \pi}\sum_{n=1}^{\infty}\frac{2^n z^{2n+1}}{(2n+1)!!}
$$
And therefore
$$
Q(a,b)=\frac{2}{\sqrt \pi}\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{-bt}\frac{2^n(i a t)^{2n+1}}{(2n+1)!!}dt=\frac{2}{\sqrt \pi}\sum_{n=1}^{\infty}\frac{(i a )^{2n+1}2^n (2n+1)!}{(2n+1)!!}=\\\frac{2 ia}{b^2\sqrt{\pi}}\sum_{n=1}^{\infty}\frac{ n! (4 a^2)^n}{(b^2)^n}
$$
using the asymptotic expansion of the Exponential integral (which is easily verified using i.p.b.)
$$
\text{Ei(z)}\sim\frac{e^{-z}}{z}\sum_{n=1}^{N-1}\frac{n!}{(-z)^n}
$$
We may ($z=-\frac{b^2}{4a^2}$) conclude that: 
$$
Q(a,b)=\frac{1}{2 a i \sqrt{\pi}}e^{\frac{b^2}{4a^2}}\text{Ei}\left(\frac{-b^2}{4a^2}\right)
$$
Q.E.D
Remark: I'm aware that the last equality sign holds only in an asymptotic way, but it seems to be possible to extend this to a real equality. If someone can hint me in the right direction i would be glad to make this point more rigouros! 
