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I'm trying to calculate the result of this integral: $$\int_{0}^{\infty} \mbox{d}x \frac{\sin(\omega x)}{x^{3/2}}$$ where $\omega > 0$, and I know the result is $\sqrt{2 \pi \omega}$, but I have no idea how to come there.

Since there is a singularity in $0$, I can't use Jordan's lemma, that could have been good since the integrand goes to $0$ uniformly for $|z| \to \infty$ in the upper half-plane ($\int_{0}^{\infty} = \frac{1+i}{2} \int_{- \infty}^{\infty}$).

I also tried to trasform the integral in a sum of integrals over every period: $$\int_{0}^{\infty} \mbox{d}x \frac{\sin(\omega x)}{x^{3/2}} = \sum_{n=0}^{\infty} \int_{2n \pi / \omega}^{2(n+1) \pi / \omega} \mbox{d}x \frac{\sin(\omega x)}{x^{3/2}} =$$ $$[z = \frac{t + 2n \pi}{\omega}, \mbox{d}z = \frac{1}{\omega} \mbox{d}t, \frac{2n \pi}{\omega} \to 0, \frac{2(n+1) \pi}{\omega} \to 2 \pi] $$ $$= \frac{\sqrt{\omega}}{2i} \sum_{n=0}^{\infty} \int_{0}^{2 \pi} \mbox{d}t \frac{e^{it} - e^{-it}}{(t + 2n \pi)^{3/2}} =$$ $$[w = e^{it}, t = -i \ln(w), \mbox{d}t = - \frac{i}{w} \mbox{d}w]$$ $$= - \frac{\sqrt{\omega}}{2} \sum_{n=0}^{\infty} \oint_{|w|=1} \mbox{d}w \frac{1 + 1/w^2}{(2n \pi - i \ln(w))^{3/2}} =$$ [Since the only singularities are in 0 (removable) and 1 (pole), for $n=0$; for $n>0$ all integrals are $=0$] $$= \sqrt{\frac{\omega}{2}} (1+i) \oint_{|w|=1} \mbox{d}w \frac{w^2 + 1}{w^2 \ln^{3/2} (w)} $$ but there is still a singularity on the path, so I can't calculate the residue and solve the integral.

I suspect that my approach is wrong somewhere...

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You've tried integrating by parts, by any chance? – J. M. Oct 29 '11 at 13:15
Yes, but with no luck. I think there is no way to calculate the primitive function. I also tried with Wolfram Aplha, the solution involves the Fresnel integral, which has to be calculated numerically if I'm not wrong. – GRB Oct 29 '11 at 13:23
So you can't use any of the methods you know on $2\omega\int_0^\infty \frac{\cos\,\omega x}{\sqrt x}\mathrm dx$? – J. M. Oct 29 '11 at 13:27
For that matter, what methods do you know for evaluating $4\omega\int_0^\infty \exp(i\omega x^2)\mathrm dx$? – J. M. Oct 29 '11 at 13:35
I see where you want to go, I'll try that way, thanks. – GRB Oct 29 '11 at 13:49
up vote 4 down vote accepted

Let's make a substitution $\omega x = y^2$ where $y=\sqrt{\omega x}$. Then $2 y \mathrm{d} y = \omega \mathrm{d} x$ : $$ \int_0^\infty \frac{\sin(\omega x)}{x^{3/2}} \mathrm{d} x = \int_0^\infty \frac{\sin(y^2)}{y^3 \omega^{-3/2}} \, \frac{2 y}{\omega} \mathrm{d} y =2 \sqrt{\omega} \int_0^\infty \frac{\sin(y^2)}{y^2} \, \mathrm{d} y = 2 \sqrt{\omega} \mathcal{I} $$ It remains to evaluate the integral $\mathcal{I}$, which is now non-singular and does not depend on $\omega$.

Integration by parts gives: $$ \begin{eqnarray} \mathcal{I} &=& \int_0^\infty \sin(y^2) \mathrm{d} \left( \frac{-1}{y} \right) = \left.-\frac{\sin(y^2)}{y} \right|_0^\infty + \int_0^\infty \frac{2 y \cos(y^2)}{y} \mathrm{d} y \\ &=& 0 + 2 \int_0^\infty \cos(y^2) \mathrm{d}y = 2 \operatorname{Re}\left( \int_0^\infty \mathrm{e}^{i y^2} \mathrm{d} y\right) = 2 \operatorname{Re}\left( \mathrm{e}^{i \pi/4} \int_0^\infty \mathrm{e}^{-y^2} \mathrm{d} y\right) = \sqrt{\frac{\pi}{2}} \end{eqnarray} $$ The last equality is obtained by rotating the integration path to ray at $\arg(y) = \frac{\pi}{4}$. Consider $\int_0^R \mathrm{e}^{i y^2} \mathrm{d} y$. Add to it the integration over the counter-clock-wise arc of radius R and the integration along the ray $\arg(z) = \frac{\pi}{4}$. The integration over so obtained closed contour is zero. Integral over arc vanishes as $R \to +\infty$. We have $\int_0^\infty \mathrm{e}^{i (\exp(i \pi/4) x)^2} \mathrm{d} (\exp(i \pi/4) x) = \exp(i \pi/4) \int_0^\infty \exp(-x^2) \mathrm{d} x = \exp(i \pi/4) \frac{\sqrt{\pi}}{2}$.

Combining results: $$ \int_0^\infty \frac{\sin(\omega x)}{x^{3/2}} \mathrm{d} x = \sqrt{2 \pi \omega} $$

It remains to show that the limit vanishes: $$ \left.-\frac{\sin(y^2)}{y} \right|_0^\infty = -\lim_{y \to \infty} \frac{\sin(y^2)}{y} + \lim_{y \to 0} \frac{\sin(y^2)}{y} $$ The first one does by sandwiching argument $-1 \le \sin(y^2) \le 1$, thus $-\frac{1}{y} \le \frac{\sin(y^2)}{y} \le \frac{1}{y}$, and both lower and upper bound have zero limit.

The latter limit is zero because $ \lim_{y \to 0} \frac{\sin(y^2)}{y} = \lim_{y \to 0} \frac{\sin(y^2)}{y^2} \cdot y = \lim_{y \to 0} \frac{\sin(y^2)}{y^2} \cdot \lim_{y \to 0} y = 1 \cdot 0$.

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The substitution $y \mbox{d}y \to \frac{1}{2} \mbox{d}y^2$ was clever, thank a lot! – GRB Oct 29 '11 at 14:00
Ok, wait, I'm not so sure about the integration by parts. Shouldn't the second part be $\int_0^{\infty} \mbox{d}y \cos(y^2)$? – GRB Oct 29 '11 at 14:31
Oops, you are right of course. Let me fix that. – Sasha Oct 29 '11 at 14:34
I think you forgot a factor 2 in the second line where you solve the I integral, and I'm not sure if you can change the argument of y like that, it seems there is a difference of a factor $\sqrt{2}$ between the two integrals: 1, 2 – GRB Oct 29 '11 at 15:07
Sasha, you might wish to add to your answer the contour integration stuff you just explained in comments, otherwise the proof is incomplete. – Did Oct 29 '11 at 16:39

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