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?


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.

  • 1
    $\begingroup$ a very nice approach! :) (+1) $\endgroup$
    – tired
    Sep 1 '15 at 14:22
  • 1
    $\begingroup$ @tired Thank you. Your approach is quite novel too. $\endgroup$
    – M.N.C.E.
    Sep 1 '15 at 14:39
  • 2
    $\begingroup$ @Chris's sis the artist Thank you. $\endgroup$
    – M.N.C.E.
    Sep 1 '15 at 14:48

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


$$ Q(a,b)=\int_0^{\infty}dte^{-b t}e^{- a^2 t^2} \text{erfc}\left(i a t\right)=\frac{1}{2 a i\sqrt{\pi}}e^{\frac{b^2}{4a^2}}\text{Ei}\left(-\frac{b^2}{4a^2}\right) $$


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:

$$ Q(a,b)=\frac{1}{2 a i \sqrt{\pi}}e^{\frac{b^2}{4a^2}}\text{Ei}\left(\frac{-b^2}{4a^2}\right) $$


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!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.