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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
    Substitute $u=\sqrt{x}$ to see that this is related to [Fresnel's integral](http://en.wikipedia.org/wiki/Fresnel_integral#Properties)...2012-07-17

6 Answers 6

19

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 $$ \begin{align} \int_0^\infty\frac{\sin(x)}{\sqrt{x}}\,\mathrm{d}x &=2\int_0^\infty\sin(u^2)\,\mathrm{d}u\\ &=2\sqrt{\frac\pi8}\\ &=\sqrt{\frac\pi2} \end{align} $$

  • 0
    would the downvoter care to comment?2014-04-06
  • 0
    I guess you take a half-circle as contour, I can't get your first equality with $\sqrt i$, could you expand please?2014-06-20
  • 0
    @kwak: I have added a section about the contour integration mentioned.2014-06-21
  • 0
    Why the second downvote?2014-06-21
  • 0
    Thanks, awesome answer, the trick is to take this quarter circle2014-06-21
  • 0
    How about after sub $u=\sqrt{x}$, we evaluate $$ 2\int_0^\infty \sin u^2\ du=2\ \Re\left[\int_0^\infty e^{-iu^2}\ du\right] $$ using Gaussian integral and get $\Re\left[\sqrt{\dfrac{\pi}{i}}\right]=\sqrt{\dfrac{\pi}{2}}$? Is that approach correct?2014-06-21
  • 0
    @Tunk-Fey: With a change of variables, that is precisely what I wrote above. However, the integrals are over two different paths and to show that they are equal, we have to use contour integration.2014-06-21
  • 0
    Truth be told, I know nothing about contour integration I never did learn that subject. I tried to justify my approach by using $\Re\left[\sqrt{\dfrac1i}\right]=\dfrac1{\sqrt{2}}$. it seems to me it's correct, my Prof in Physics also felt OK with that but I wonder how does this approach matter in mathematician's viewpoint?2014-06-22
  • 0
    @Tunk-Fey contour integration comes with the study of holomorphic functions (functions continuous in $\Bbb C$)2014-06-27
12

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.

11

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.

8

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

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

6

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.

  • 0
    Is it **obvious** what the result of the last indefinite integral is?2012-07-17
  • 0
    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.2012-07-18
  • 0
    @J. M.: agree. My answer was updated.2012-07-18
0

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 $$ \int_0^\infty \frac{t^p}{t^2+1}dt=\frac{\pi}{2\cos\frac{p\pi}{2}}, |p|<1. $$