How does one compute the following integral? $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$

I have tried extending $x$ to the complex plane then evaluating the following contour integral $$\oint_C \frac{\sqrt{x}e^{ix}}{1+x^2} dx$$ with the contour $C$ running along the whole real axis and then upper semicircle. I obtain $$\int_{0}^{\infty} \frac{\sqrt{x}\big(\cos(x)+\sin(x)\big)}{1+x^2}\,\mathrm dx=\frac\pi{e\sqrt2}$$ but not the original integral.

  • $\begingroup$ WA does not appear to give the solution in terms of regular constants. It suggests a need to use error functions and the substitution $x\mapsto x^2$ followed by partial fractions on the denominator. $\endgroup$ – Simply Beautiful Art Aug 12 '17 at 0:21
  • $\begingroup$ @SimplyBeautifulArt: The solution does not have to be regular constants, so long it is in special functions. I thought of that substitution before. How then do you deal with $\sin(x^2)$? $\endgroup$ – Hans Aug 12 '17 at 0:29
  • $\begingroup$ @SimplyBeautifulArt: I think I know what to do. Do use the transformation then pick the contour to be around the first quadrant of the complex plane and take care of the singularity. I will try this later --- occupied right now. $\endgroup$ – Hans Aug 12 '17 at 0:37
  • $\begingroup$ Yeah. And replace the sine with an exponential function (which is where we get our ole friend error function) $\endgroup$ – Simply Beautiful Art Aug 12 '17 at 0:39
  • 2
    $\begingroup$ @Hans: After a simple substitution, we are ultimately left with evaluating $J(i),~$ where $$J(a)~=~\int_0^\infty\frac{e^{-ax^2}}{x^2+x^{-2}}~dx,$$ which is nothing more than $-F'(a),~$ for $$F(a) ~=~ \int_0^\infty\frac{e^{-ax^2}}{x^4+1}~dx,$$ which, in its turn, constitutes a solution to the differential functional equation $$F(a)+F''(a)=\dfrac12~\sqrt{\dfrac\pi a},$$ see Gaussian integral for more details. $\endgroup$ – Lucian Aug 30 '17 at 4:42

For information :

$\int_{0}^{\infty} \frac{\sqrt{x}\sin(y\:x)}{1+x^2} dx$ is the Fourier Sine Transform of $\frac{\sqrt{x}}{1+x^2}\quad $ In the Harry Bateman's Tables of Integral Transforms an even more general formula can be seen on page 71, Eq.28 , the Fourier Sine Transform of $x^{2\nu}(x^2+a^2)^{-\mu-1}$ : $$\frac{1}{2}a^{2\nu-2\mu}\frac{\Gamma(1+\nu)\Gamma(\mu-\nu)}{\Gamma(\mu+1)}y \:_1\text{F}_2(\nu+1;\nu+1-\mu,3/2;a^2y^2/4)\:+\:4^{\nu-\mu-1}\sqrt{\pi}\frac{\Gamma(\nu-\mu)}{\Gamma(\mu-\nu+3/2)}y^{2\mu-2\nu+1}\:_1\text{F}_2(\mu+1;\mu-\nu+3/2,\mu-\nu+1;a^2y^2/4) $$ With$\quad y=1\quad;\quad a=1\quad;\quad \nu=1/4\quad;\quad \mu=0\quad\to\quad \int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =$
$$=\frac{1}{2}\Gamma(5/4)\Gamma(-1/4) \:_1\text{F}_2(5/4;5/4,3/2;1/4)\:+\:4^{-3/4}\sqrt{\pi}\frac{\Gamma(5/4)}{\Gamma(5/4)}\:_1\text{F}_2(1;5/4,3/4;1/4) $$ The Generalized Hypergeometric $_1$F$_2$ function (don't confuse with the well-known 2F1) reduces to functions of lower level in the particular cases :


$\:_1\text{F}_2(1;5/4,3/4;1/4)=\frac{\sqrt{\pi}}{4e}\left(e^2\text{erfi}(1)+\text{erf}(1) \right)$

and after simplification : $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx = -\frac{\pi}{\sqrt{2}}\sinh(1)+\frac{\pi}{2\sqrt{2}\:e}\left(e^2\text{erfi}(1)+\text{erf}(1) \right)$$ $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(e^2\text{erfi}(1)+\text{erf}(1) +1-e^2\right)$$ $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right)$$ For the Hypergeometric $_1$F$_2$ function, see : http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F2/02/

About the functions erf, erfi, erfc, see : http://mathworld.wolfram.com/Erf.html , http://mathworld.wolfram.com/Erfi.html , http://mathworld.wolfram.com/Erfc.html

  • $\begingroup$ Nice! +1. Where can I find the derivation of the transforms, at least this particular one, in Harry Bateman's Tables of Integral Transforms? $\endgroup$ – Hans Aug 31 '17 at 19:52
  • $\begingroup$ The Bateman's Tables of Integral Transforms (1954), is a extensive compilation of equations of integral transforms, without reference. Sorry, I have no more information. $\endgroup$ – JJacquelin Aug 31 '17 at 20:18

This doesn't answer your specific question, but I thought it might be worth mentioning that we can generalize the result you found using the same contour.

Namely, $$\int_{0}^{\infty} \frac{x^{s-1} \sin \left(x-\frac{\pi s}{2} \right)}{1+x^{2}} \, dx = -\frac{\pi}{2e},\quad 0 <s <3. $$

This integral (with a couple more parameters) appears as an exercise on page 154 in the textbook An Introduction to the Theory of Functions of a Complex Variable by E.T. Copson. Copson attributes it to Cauchy.

What makes this integral somewhat interesting is the fact that its value is independent of $s$.

By integrating $$f(z) =z^{s-1} \, \frac{e^{- i \pi s/2}e^{iz}}{1+z^{2}} $$ around a large semicircular contour in the upper half-plane that is indented at the origin and applying Jordan's lemma, we get

$$ \begin{align} \int_{-\infty}^{0} (|x|e^{i \pi})^{s-1} \, \frac{e^{- i \pi s/2} e^{ix}}{1+x^{2}} \, dx + \int_{0}^{\infty} x^{s-1} \frac{e^{- i \pi s/2}e^{ix}}{1+x^{2}} \, dx &= 2 \pi i \operatorname{Res} [f(z), e^{i \pi /2}] \\ &= 2 \pi i \left( (e^{i \pi /2})^{s-1} \frac{e^{- i \pi s/2} e^{-1}}{2i}\right)\\ & = \frac{\pi}{i e}. \end{align}$$

(As long as $s>0$, the contribution from the indentation around the origin will vanish as the radius of the indentation goes to $0$.)

But the left side of the equation can be written as $$-\int_{0}^{\infty} u^{s-1} \, \frac{e^{i \pi s/2}e^{-iu}}{1+u^{2}} \, du+ \int_{0}^{\infty} x^{s-1} \frac{e^{- i \pi s/2}e^{ix}}{1+x^{2}} \, dx = 2i \int_{0}^{\infty} x^{s-1} \, \frac{\sin\left(x- \frac{\pi s}{2} \right)}{1+x^{2}} \, dx, $$ and the result follows.

I originally posted something similar as an answer to a different question, but I think it's more appropriate to post it here.

  • $\begingroup$ This is exactly the contour and approach I described I was trying in my question. The integral I obtained is precisely the one you quoted when $s=\frac32$. The intrigue of the independence of the integral from $s$ is alleviated by the fact that had it been dependent there is a multiplicative factor and dividing that factor would have made the right hand side a constant. But this general integral is interesting in its own right. $\endgroup$ – Hans Nov 27 '17 at 3:21
  • $\begingroup$ Right before you posted your comment, I added the words "more generally" to my answer to reflect the fact that I was generalizing the result you got. $\endgroup$ – Random Variable Nov 27 '17 at 3:53
  • $\begingroup$ I see. I do think it is interesting. That was why I upvoted your answer. It would be better if you could explicitly say this is a generalization of my result even though it does not answer the original question per se. $\endgroup$ – Hans Nov 27 '17 at 6:24

My goal is to prove $$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right)$$ as stated in @JJacquelin's answer. My following attempt falls short of that. But it does provide an equivalent integral with much faster convergence.

Take the contour integral $$2\oint_C \frac{z^2e^{iz^2}}{1+z^4} dz$$ with contour $C$ run along the real axis from $0$ to $R$, trace the circle $Re^{i\theta}$ with $\theta$ running from $0$ to $\frac\pi 4$, then roll back to the origin along $xe^{i\frac\pi 4}$ with $x$ running from $R$ to $0$ while circumventing around the point $e^{i\frac\pi 4}$ clockwise with radius $\delta>0$. The integral on the octal circle vanishes as $R\to\infty$. We have $$\oint_C \frac{z^2e^{iz^2}}{1+z^4} dz=\int_{0}^{\infty} \frac{x^2e^{ix^2}}{1+x^4} dx-e^{i\frac34\pi}\int_0^\infty \frac{x^2e^{-x^2}}{1-x^4}dx+\cdots$$

(to be continued)


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.