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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?

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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.

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    $\begingroup$ a very nice approach! :) (+1) $\endgroup$ – tired Sep 1 '15 at 14:22
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    $\begingroup$ @tired Thank you. Your approach is quite novel too. $\endgroup$ – M.N.C.E. Sep 1 '15 at 14:39
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    $\begingroup$ @Chris's sis the artist Thank you. $\endgroup$ – M.N.C.E. Sep 1 '15 at 14:48
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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!

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    $\begingroup$ Don't be lazy!:-) Good job! (+1) $\endgroup$ – user 1357113 Sep 1 '15 at 14:46

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