$\int_0^{\infty} \frac{\sin x}{x^p} dx$

Question

Evaluate$$\int_0^{\infty} \frac{\sin x}{x^p} dx$$

I tried to use same trick as we do to evaluate ${sinx\over x}$ . But i couldn't find a suitable partial function to differentiate and then integrate . Also tried writing it as $\Im[ \int_0^{\infty} \frac{e^{ix}}{x^p} dx]$ and use gamma function but powers of iota is confusing me .

I don't want to consider convergence of function here since there are already many similar questions . So p is such that integral very well converges .

@Mostafa Ayaz, I prefer \operatorname{Im} because it produces spacing as operator should. You can compare \operatorname{Im}f and \text{Im}f. — Ennar, Jan 03 '22 at 17:39
https://math.stackexchange.com/questions/1222832/how-to-integrate-int-0-infty-sin-x-cdot-x-1-3-dx-using-gamma-functio — Svyatoslav, Jan 03 '22 at 17:51

Stefan Lafon · Accepted Answer · 2022-01-03T17:59:22.357

4

$$\begin{split} \int_0^{+\infty}\frac{\sin x}{x^p}dx &= \int_0^{+\infty}\frac{\sin x }{\Gamma(p)}\int_0^{+\infty}t^{p-1}e^{-xt}dtdx\\ &= \int_0^{+\infty}\frac{t^{p-1}}{\Gamma(p)}\int_0^{+\infty}\sin(x)e^{-xt}dxdt\\ &= \frac 1 {\Gamma(p)}\int_0^{+\infty}\frac{t^{p-1}}{1+t^2}dt\\ &= \frac 1 {\Gamma(p)}\frac{\pi}{2\sin \left(\frac {p\pi} 2\right) } \end{split}$$ where the last equality comes from this answer.

edited Jan 03 '22 at 17:59

answered Jan 03 '22 at 17:52

Stefan Lafon

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I think you need $p<2$ as otherwise the integral diverges at $0$. – Rene Schipperus Jan 03 '22 at 18:58
Why writing sinx as $\Im e^{ix}$ doesn't work here it seems quite simple ? – RKK Jan 03 '22 at 18:59
@ReneSchipperus That's right, but the OP said "$p$ is such that integral very well converges", so I was assuming $p$ was in the right range. In fact I was assuming $0<p<1$ to make the original integral converge. It's interesting that you can extend the right-hand side to all $p\notin 2\mathbb N$. – Stefan Lafon Jan 03 '22 at 18:59
@RKK Correct, using the imaginary part is not helpful in this case. I wouldn't personally think the answer is simple, because to show the last equality, you need to a bit of work with Fourier series. – Stefan Lafon Jan 03 '22 at 19:01
I suspect that Answer has something to do with Euler's reflection formula ? – RKK Jan 03 '22 at 19:08
You are correct. One way, for instance, to show the last equality is to use the reflection formula, as in https://proofwiki.org/wiki/Euler%27s_Reflection_Formula – Stefan Lafon Jan 03 '22 at 19:12
In a nbd of $0$ $\sin x$ is like $x$ so the integral is basically $\int_0 \frac{1}{x^{p-1}}dx$ which by direct integration converges for $p-1<1$. – Rene Schipperus Jan 03 '22 at 19:50
Ooops you're absolutely right. $p<2$. – Stefan Lafon Jan 03 '22 at 20:02

score 1 · Answer 2 · answered Jan 03 '22 at 20:00

The given integral is dependent on the value of $p$.

Taking a look at my formula collection I find a brand new formula showing up that the representation of the infinite integral is still under research.

$\int_{0}^{\infty}\frac{sin(x)}{x^{p}}=x^{-p}((-ix)^{p}\Gamma(-p,-ix)+(ix)^{p}\Gamma(1-p,-ix) + constant$

$\Gamma(a,b)$ is the incomplete gamma function.

$\int_{0}^{\infty}\frac{sin(x)}{x^{p}}=\frac{i x^{1-p}}{2}(E_{p}(-ix)+E_{p}(ix)) + constant $

$E_{n}(x)$ is the exponential integral $E$.

Where $p$ can be arbitrary complex.

It is possible to get an easier representation for $2>\Re(p)>0$. This is

$\int_{0}^{\infty}\frac{sin(x)}{x^{p}}=cos(\frac{p \pi}{2})\Gamma(1-p)$

That is rather different from the already given solution. Prove this under the assumption that the formulas are valid and the main theorem of the integral and differential calculus. This is simply the definition of the $\Gamma(a,b)$ is the incomplete gamma function.

Look for example into this answers: definition of incomplete gamma function.

Wolfram Alpha based on version 13 is showing this:

$\int_0^{\infty} \frac{\sin x}{x^p} dx$

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