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For $r \in (0,2)$, I would like to evaluate the integral

$$\frac{2}{r} \int_0^{\infty} \frac{\sin(u)}{u^r} du.$$

The answer should be

$$\frac{\pi \cdot \mathrm{cosec}{\frac{r\pi}{2}} }{\Gamma(r+1)}. $$

Many thanks for your help.

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  • $\begingroup$ your recent edit corrected an error in your typesetting. But that is not why your question is on hold. As the notice says, give some additional context, especially on the work you have done so far on this problem. $\endgroup$ – Rory Daulton Apr 22 '15 at 18:32
  • $\begingroup$ This makes no sense. It's a simple, well-posed question, now with three answers. It's an integral I wanted to compute. That's the context - nothing more. $\endgroup$ – Frank Apr 22 '15 at 19:41
  • $\begingroup$ @Frank: people here don't like no-effort, homework questions. Unfortunately, many people put questions like this in that category. I strongly disagree, but there you have it. $\endgroup$ – Ron Gordon Apr 22 '15 at 20:33
  • $\begingroup$ Sure, I understand that - but, for the record, this isn't homework, just something I came across in my reading, and had no idea how to solve. $\endgroup$ – Frank Apr 22 '15 at 20:38
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Write

$$\int_0^{\infty} du \frac{\sin{u}}{u^r} = \operatorname{Im}{\left [ PV \int_0^{\infty} du \, u^{-r} e^{i u} \right ]} $$

Now consider the contour integral

$$\oint_C dz \, z^{-r} e^{i z} $$

where $C$ is a 90-degree circular wedge of radius $R$ in the first quadrant of the complex plane, with a quarter circle of radius $\epsilon$ cut out at the origin. Then the contour integral is equal to

$$\int_{\epsilon}^R dx \, x^{-r} \, e^{i x} + i R \int_0^{\pi/2} d\theta\, e^{i \theta} R^{-r} \, e^{-i r \theta} e^{i R e^{i \theta}} \\ + i \int_R^{\epsilon} dy \, e^{-i \pi r/2} y^{-r} e^{-y} + i \epsilon \int_{\pi/2}^0 d\theta\, e^{i \theta} \epsilon ^{-r} \, e^{-i r \theta} e^{i \epsilon e^{i \theta}}$$

The second integral vanishes as $R \to \infty$ because its magnitude is bounded by

$$R^{-(r-1)} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le R^{-(r-1)} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} \le \frac{\pi}{2 R^{r}} $$

which vanishes as $r \gt 0$. Now, as $\epsilon \to 0$, the fourth integral behaves as

$$i \epsilon^{1-r} \int_{\pi/2}^0 d\theta \, e^{-i (r-1) \theta} \left [ 1+ O(\epsilon) \right ] = \frac{\epsilon^{1-r}}{1-r} \left ( 1-i e^{-i \pi r/2}\right ) + O(\epsilon^{2-r}) $$

There are two cases.

1) When $r \gt 1$, there is an apparent singularity at the origin, but this will cancel out with another from the other integrals.

Now, rewrite

$$\int_{\epsilon}^{\infty} dx \, x^{-r} \, e^{i x} = \int_{\epsilon}^\infty dx \, x^{-r} \, (e^{i x}-1) + \int_{\epsilon}^\infty dx \, x^{-r} = \int_{\epsilon}^\infty dx \, x^{-r} \, (e^{i x}-1) - \frac{\epsilon^{1-r}}{1-r}$$

Similarly,

$$i \int_{\infty}^{\epsilon} dy \, e^{-i \pi r/2} y^{-r} e^{-y} = i \int_{\infty}^{\epsilon} dy \, e^{-i \pi r/2} y^{-r} (e^{-y}-1) + i e^{-i \pi r/2} \frac{\epsilon^{1-r}}{1-r} $$

Note that the sum of the two integrated pieces cancels out the singular piece from the fourth integral. The remaining integrals indeed converge as $\epsilon \to 0$ and we have that the contour integral is finally equal to

$$ -\int_{0}^\infty dx \, x^{-r} \, (1-e^{i x}) + i e^{-i \pi/2} \int_{0}^\infty dy \, y^{-r} (1-e^{-y}) $$

By Cauchy's theorem, the contour integral is zero. Thus, taking imaginary parts, we finally have

$$\begin{align}\int_{0}^\infty dx \, x^{-r} \sin{x} &= - \cos{\left ( \frac{\pi r}{2}\right )} \int_{0}^\infty dy \, y^{-r} (1-e^{-y})\\ &= - \cos{\left ( \frac{\pi r}{2}\right )} \int_{0}^{\infty} dy \, y^{1-r} \, \int_0^1 du \, e^{-y u}\\ &= - \cos{\left ( \frac{\pi r}{2}\right )} \int_0^1 du \, \int_0^{\infty} dy \, y^{1-r} e^{-y u} \\ &= - \cos{\left ( \frac{\pi r}{2}\right )} \Gamma(2-r) \int_0^1 du \, u^{r-2} \\ &= \cos{\left ( \frac{\pi r}{2}\right )} \Gamma(1-r) \end{align}$$

2) When $r \lt 1$, there is no singular piece. Thus, by Cauchy's theorem, and taking imaginary parts, we may write that

$$\int_{0}^\infty dx \, x^{-r} \sin{x} = \cos{\left ( \frac{\pi r}{2}\right )} \int_0^{\infty} dy \, y^{-r} e^{-y} = \cos{\left ( \frac{\pi r}{2}\right )} \Gamma(1-r) $$

The answer is the same in both cases. When we apply the reflection formula, we find that

$$\cos{\left ( \frac{\pi r}{2}\right )} \Gamma(1-r) = \frac{\pi \cos{\left ( \frac{\pi r}{2}\right )}}{\sin{(\pi r)}\, \Gamma(r)} = \frac{\pi r}{ 2 \sin{\left ( \frac{\pi r}{2}\right )} \Gamma(1+r)}$$

$$\frac{2}{r} \int_{0}^\infty dx \, x^{-r} \sin{x} = \frac{\pi}{ \sin{\left ( \frac{\pi r}{2}\right )} \Gamma(1+r)} $$

as was sought.

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Hint: Use Euler's formula in conjunction with the well-known integral expression for the $\Gamma$ function, and then employ the reflection formula.

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Write $$ \frac{1}{ru^r} = \frac{1}{\Gamma(r+1)}\int_0^{\infty} \alpha^{r-1} e^{-u\alpha} \, d\alpha. $$ Then your integral becomes, after swapping the order of integration, $$ \frac{2}{\Gamma(r+1)} \int_0^{\infty} \alpha^{r-1} \left( \int_0^{\infty} e^{-u\alpha} \sin{u} \, du \right) \, d\alpha. $$

Now, we can do the internal integral by integration by parts or using the complex expansion of the sine. Either way, the inside integral is $\frac{1}{1+\alpha^2}$, so we're down to $$ \frac{2}{\Gamma(r+1)} \int_0^{\infty} \frac{\alpha^{r-1}}{1+\alpha^2} \, d\alpha $$ Change variables, $u=\alpha^2$, $du/u = 2\, d\alpha/\alpha$ and this turns into $$ \frac{1}{\Gamma(r+1)}\int_0^{\infty} \frac{u^{r/2-1}}{1+u} \, du $$ The remaining integral is well-known (turn it into the beta function, or do this), and gives the cosecant, so the final answer is $$ \frac{\pi \operatorname{cosec}{(\pi r/2)}}{\Gamma(r+1)} $$


And curiously, the way the question is written, either this or Ron Gordon's answer could be what is wanted...

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  • $\begingroup$ Hi - thanks a lot for your answer. I don't understand what is going on with this question though. Why is does the way the question is written make anything curious? Your answer and Ron's are the same. $\endgroup$ – Frank Apr 22 '15 at 18:28
  • $\begingroup$ Although the results are the same, the methods are rather different: @RonGordon's uses contour integration and a wedge-shaped contour to turn the integrand into one that looks like the Gamma-function, whereas mine uses the scaling properties of the Gamma-function integral to rewrite the integral as a double integral, change the order of integration and then do the easier integrals that result. My remark is that since both of these use the Gamma-function (in different ways), either is a reasonable way of answering the question (although mine is probably not the "usual" way). $\endgroup$ – Chappers Apr 23 '15 at 9:11

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