Numerically it seems to be true that

$$ \int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}. $$

Any ideas how to prove this?

  • 3
    $\begingroup$ Substitute $u=\sqrt{x}$ to see that this is related to Fresnel's integral... $\endgroup$ – draks ... Jul 17 '12 at 15:24

Using contour integration, we get $$ \begin{align} \int_0^\infty\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x &=\sqrt{i\,}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x\\ &=\frac{1+i}{\sqrt{2}}\Gamma\left(\frac12\right)\\ &=(1+i)\sqrt{\frac\pi2} \end{align} $$ Therefore, $$ \int_0^\infty\frac{\cos(x)}{\sqrt{x}}\,\mathrm{d}x=\int_0^\infty\frac{\sin(x)}{\sqrt{x}}\,\mathrm{d}x=\sqrt{\frac\pi2} $$

About the Contour Integration

If we integrate $f(z)=\dfrac{e^{iz}}{\sqrt{z}}$ around the contour $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$ as $R\to\infty$, we get that $$ \int_0^R\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x +\int_0^{\pi/2}\frac{e^{iRe^{ix}}}{\sqrt{R}e^{ix/2}}iRe^{ix}\,\mathrm{d}x -\sqrt{i\,}\int_0^R\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x =0 $$ because there are no singularities of $f$ inside the contour. Then because $$ \begin{align} \left|\int_0^{\pi/2}\frac{e^{iRe^{ix}}}{\sqrt{R}e^{ix/2}}iRe^{ix}\,\mathrm{d}x\right| &\le\sqrt{R}\int_0^{\pi/2}e^{-R\sin(x)}\,\mathrm{d}x\\ &\le\sqrt{R}\int_0^{\pi/2}e^{-2Rx/\pi}\,\mathrm{d}x\\ &\le\frac\pi{2\sqrt{R}} \end{align} $$ vanishes as $R\to\infty$, we have $$ \int_0^\infty\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x =\sqrt{i\,}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x $$

Real Method

Substituting $u^2=x$ and applying this answer, which uses only real methods, yields $$ \begin{align} \int_0^\infty\frac{\sin(x)}{\sqrt{x}}\,\mathrm{d}x &=2\int_0^\infty\sin(u^2)\,\mathrm{d}u\\ &=2\sqrt{\frac\pi8}\\ &=\sqrt{\frac\pi2} \end{align} $$

  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Apr 6 '14 at 9:41
  • $\begingroup$ I guess you take a half-circle as contour, I can't get your first equality with $\sqrt i$, could you expand please? $\endgroup$ – caub Jun 20 '14 at 23:10
  • $\begingroup$ @kwak: I have added a section about the contour integration mentioned. $\endgroup$ – robjohn Jun 21 '14 at 1:01
  • $\begingroup$ Why the second downvote? $\endgroup$ – robjohn Jun 21 '14 at 1:14
  • $\begingroup$ Thanks, awesome answer, the trick is to take this quarter circle $\endgroup$ – caub Jun 21 '14 at 9:11

That is a Fresnel integral.

Make the substitution $\sqrt{x}=u$. Then you get $dx=2udu$ from where

$$\int_0^\infty \sin(x) x^{-1/2} dx =2 \int_0^\infty \sin(u^2)du=2\sqrt{\frac{\pi}{8}}=\sqrt{\frac{\pi}{2}}$$

See this question for more details.


Here is an another approach, which I show only a heuristic calculation:

$$\begin{align*} \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} \; dx &= \int_{0}^{\infty} \left( \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} t^{-1/2} e^{-xt} \; dt \right) \sin x \; dx \\ &\stackrel{\ast}{=} \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} t^{-1/2} \int_{0}^{\infty} e^{-xt} \sin x \; dx \; dt \\ &= \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} \frac{t^{-1/2}}{1+t^2} \; dt \\ &= \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} \frac{2du}{1+u^4}.\qquad(t = u^2) \end{align*}$$

Now it is not hard to show that

$$ \int_{0}^{\infty} \frac{2du}{1+u^4} = \frac{\pi}{\sqrt{2}}.$$

Indeed, you may use the equality

$$ \begin{align*} \frac{2 du}{1+u^4} &= \frac{2u^{-2} \; du}{u^2 + u^{-2}} = \frac{1 + u^{-2} \; du}{u^2 + u^{-2}} - \frac{1 - u^{-2} \; du}{u^2 + u^{-2}}\\ &= \frac{d\left(u - u^{-1}\right)}{\left(u - u^{-1}\right)^2 + 2} - \frac{d\left(u + u^{-1}\right)}{\left(u + u^{-1}\right)^2 - 2} \end{align*}$$

and hence deduce that

$$ \begin{align*} \int_{0}^{\infty} \frac{2 du}{1+u^4} &= \int_{0}^{\infty} \frac{d\left(u - u^{-1}\right)}{\left(u - u^{-1}\right)^2 + 2} - \int_{0}^{\infty} \frac{d\left(u + u^{-1}\right)}{\left(u + u^{-1}\right)^2 - 2}\\ &= \int_{-\infty}^{\infty} \frac{dv}{v^2 + 2} - \color{blue}{\int_{\infty}^{\infty} \frac{dw}{w^2 - 2}} \qquad \begin{pmatrix}v = u - u^{-1} \\ w = u + u^{-1}\end{pmatrix}\\ &= \frac{\pi}{\sqrt{2}} + 0. \end{align*}$$

as claimed, where the blue-colored integration is taken along the curve starting from $+\infty$ to $2$, and then turning back to $+\infty$, which makes it cancel out. Therefore we obtain the desired result.

The problem in this calculation is that the starred equality is almost unable to be justified by any simple means.


From the result

$$ \int_0^\infty dx \frac{\sin x}{\sqrt x} = 2 \int_0^\infty dx \sin\left(x^2\right), $$

I'd use $\exp\left(-i x^2\right) = \cos \left(x^2\right) - i \sin \left(x^2\right)$ and

$$ \int_0^\infty dx \ \exp\left(- a x^2\right) = \frac{1}{2} \sqrt{\frac{\pi}{a}}. $$


Let's start out with the following relation: $$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx = \frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+(a+x^2)^2} dx \tag1$$ Proof of the relation $(1)$

$$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx=$$ Notice that $\displaystyle \frac{1}{\sqrt x}= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-xt^2} dt$ and have that

$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \sin x e^{-ax}\left(\int_{0}^{\infty} e^{-xt^2} dt\right) dx=$$ $$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dt\right) dx=$$ Change the integration order $$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx\right) dt=$$ Now let's recollect the formula $$ \int e^{\alpha x} \sin (\beta x) \ dx = \frac{e^{\alpha x}(-\beta (\cos (\beta x) + \alpha \sin(\beta x)))}{{\alpha}^2+{\beta}^2}$$

Hence $$\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx=-\frac{e^{-(a+t^2)x}((a+t^2)\sin x + \cos x)}{1+(a+t^2)^2}\bigg|_{0}^{\infty}=\frac{1}{1+(a+t^2)^2}$$ Then $$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx\right) dt=\frac{2}{\sqrt{\pi}} \int_{0}^{\infty}\frac{1}{1+(a+t^2)^2} \ dt.$$ End of the relation $(1)$ proof.

Based upon the above relation we get that $$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} dx=$$ $$ \lim_{a\to0+} \int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx =$$ $$ \lim_{a\to0+} \frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+(a+x^2)^2} dx=$$ $$\frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+x^4} dx \tag2$$ For the last integral we may change the variable and everything gets reduced to computing beta function

Change the variable $$x=\left(\frac{t}{1-t}\right)^{\frac{1}{4}}$$ Then $$\int_0^\infty \frac{1}{1+x^4} \ dx = \int_0^1 \frac{1}{4} (1-t)^{\frac{3}{4}-1} t^{\frac{1}{4}-1} \mathrm{d} t = \frac{1}{4} \operatorname{B}\left(\frac{1}{4}, \frac{3}{4}\right) = \frac{1}{4} \sqrt{2} \pi \tag3$$

Finally, from $(2)$ and $(3)$ we obtain the desired result

$$\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}.$$ Q.E.D.

  • $\begingroup$ Is it obvious what the result of the last indefinite integral is? $\endgroup$ – Pedro Tamaroff Jul 17 '12 at 19:15
  • $\begingroup$ Still, don't write "obviously" when you mean to say that somebody already did the work for you; it's better to link to the answer you refer to instead. $\endgroup$ – J. M. ain't a mathematician Jul 18 '12 at 11:21
  • $\begingroup$ @J. M.: agree. My answer was updated. $\endgroup$ – user 1591719 Jul 18 '12 at 11:33

Here I use Laplace Transform to present a simple proof and do not need use other tools. Note that $$ \int_0^\infty e^{-xt}\frac{1}{\sqrt{\pi t}}dt=\frac{1}{\sqrt x} $$ and hence \begin{eqnarray} \int_0^\infty\frac{\sin x}{\sqrt x}dx&=&\int_0^\infty\sin x\left(\int_0^\infty e^{-xt}\frac{1}{\sqrt{\pi t}}dt\right)dx\\ &=&\frac{1}{\sqrt\pi}\int_0^\infty \left(\int_0^\infty e^{-xt}\sin xdx\right) \frac{1}{\sqrt{t}}dt\\ &=&\frac{1}{\sqrt\pi}\int_0^\infty \frac{1}{t^2+1} \frac{1}{\sqrt{t}}dt\\ &=&\frac{1}{\sqrt\pi}\int_0^\infty \frac{\sqrt t}{t^2+1}dt\\ &=&\frac{1}{\sqrt\pi}\frac{\pi}{\sqrt 2}\\ &=&\sqrt{\frac{\pi}{2}}. \end{eqnarray} Here we use the following well-known integral $$ \int_0^\infty \frac{t^p}{t^2+1}dt=\frac{\pi}{2\cos\frac{p\pi}{2}}, |p|<1. $$


The great solution using the integration contour leads to an extended result (not only for $s=1/2$): $$\int\limits_0^\infty x^{s-1} \sin x \,\mathrm{d}x = \Gamma(s)\sin\frac{s\pi}2 \quad (s\in\mathbb{R}; 0<s<1)$$ If we integrate $f(z)=z^{s-1}e^{iz}$ around the contour $[0,R]\cup R e^{i[0,\pi/2]}\cup i[R,0]$ as $R\to\infty$, and because there are no singularities of $f$ inside the contour, the Cauchy theorem gives: $$\int\limits_0^R x^{s-1} e^{ix} \,\mathrm{d} x + \int\limits_0^{\pi/2} \underbrace{R^{s-1} e^{i\theta(s-1)} e^{iRe^{i\theta}} Ri}_{g(R,\theta)} \,\mathrm{d} \theta - i^s \int\limits_0^R y^{s-1} e^{-y} \,\mathrm{d} y =0.$$ The second term vanishes as $R\to\infty$ since \begin{align*} \left| \int_0^{\pi/2} g(R,\theta) \,\mathrm{d}\theta \right| & \le \int_0^{\pi/2} \left| g(R,\theta)\right| \,\mathrm{d}\theta \\ & \le R^s \int_0^{\pi/2} e^{-R\sin\theta} \,\mathrm{d}\theta \le R^s \int_0^{\pi/2} e^{-2R\theta/\pi} \,\mathrm{d} \theta \\ & \le R^{s-1} \frac{\pi}2 \quad \to \quad 0. \end{align*} so we have $$\int\limits_0^\infty x^{s-1} e^{ix} \,\mathrm{d} x = e^{is\pi/2} \ \Gamma(s),$$ which imaginary part is the result.

  • $\begingroup$ The function $f(z)$ has a singularity at $0$, so can't consider that contour. $\endgroup$ – jjagmath Nov 26 '19 at 1:47

The people who are contour integrating need to account for the branch point that occurs at x = 0. The quarter circle works fine, but you need to indent the contour around z = 0.

$\oint_{C}= \oint_{C_{1}}+\oint_{C_{2}}+\oint_{C_{3}}+\oint_{C_{4}}$

$C_{1}:z=x ,z \in[\epsilon,R]$

$C_{2}:z=Re^{i\theta} ,\theta \in[0,\frac{\pi}{2}]$

$C_{3}:z=iy ,\theta \in[R,\epsilon]$

$C_{4}:z=\epsilon e^{i\theta} ,\theta \in[\frac{\pi}{2},0]$

You need to estimate the fourth integral to be zero, which isn't nontrivial.

$\oint_{C_{4}}\frac{exp(iz)}{\sqrt{z}}= -i\int_0^\frac{\pi}{2}\frac{e^{i\epsilon e^{i\theta}}}{\sqrt{\epsilon e^{i\theta}}}\epsilon e^{i \theta}d\theta$

$-i\int_0^\frac{\pi}{2}e^{i\epsilon e^{i\theta}}\sqrt{\epsilon} e^{\frac{i\theta}{2}} d\theta$

You can expand the main exponential in power series about $\epsilon = 0$


You will be integrating a function that is power series in $\epsilon$ and $e^{i\theta}$, a term by term integration over those will lead to integrals that converge. You can then take the limit as $\epsilon \longrightarrow 0$, which shows that the fourth integral equals zero.


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