Evaluate the integral $$\int_0^\infty \dfrac{\sin(ax)}{x^b} \, dx,\quad a\in \mathbb{R},\quad 0<b<2.$$
I know $a=1$ ,and $ b\in \mathbb{N}$, I can find the value,
How to evaluate this integral for general $a,b$? Thank you.
Asked
Active
Viewed 568 times
6
-
The indefinite integral is sure to be nasty based on what http://integral-table.com/integral-table.html has for similar integrals. Maybe that improper integral will magically be nice somehow, though. – dfeuer Jul 29 '13 at 05:37
-
For $a\neq 0$ you may substitute $u=ax$; hence it suffices to solve the problem for $a=1$. – vadim123 Jul 29 '13 at 05:49
-
Yes,But $b\in R,0<b<2$ – math110 Jul 29 '13 at 05:49
-
Wolfram gives $\int_0^\infty \frac{\sin x}{x^b}dx=\cos(\pi b/2)\Gamma(1-b)$. – vadim123 Jul 29 '13 at 05:54
-
Maple produces $${a}^{b-1}\Gamma \left( 1-b \right) \cos \left( 1/2,\pi ,b \right) $$ for $a>0,,0<b,,b<2.$ – user64494 Jul 29 '13 at 07:02
-
Here is a related problem. – Mhenni Benghorbal Jul 29 '13 at 09:46
2 Answers
9
Process 1:$$\int_0^\infty \dfrac{\sin(ax)}{x^b} \, dx.$$$$=-\Im\left(\int_0^\infty\large e^{-iax}x^{-b}\, dx\right)$$$$=-\Im\left(\frac{1}{(ia)^{1-b}}\int_0^\infty\large e^{-t}t^{-b}\, dt\right)$$$$=-\Im\left(\frac{1}{(ia)^{1-b}}\Gamma(1-b)\right)=a^{b-1}\Gamma(1-b)\cos \left({\pi b\over 2}\right)$$
Process 2: Use Mellin Transformation.
Let, $$I(a,b)=\int_0^\infty \dfrac{\sin(ax)}{x^b} \, dx=a^{b-1}\int_0^\infty \sin(x)x^{-b}\,dx$$ Now, using Mellin transformation of $\sin x$, we will have, $$I(a,b)=a^{b-1}\Gamma(1-b)\sin\left(\frac{\pi(1-b)}{2}\right)=a^{b-1}\Gamma(1-b)\cos \left({\pi b\over 2}\right)$$
Kunnysan
- 2,050
-
I think there is a minus sign issue in front of the integral on the 3rd line. – user 1591719 Jul 29 '13 at 09:34
-
-
1
Note that $\sin x$ can be decomposed into the exponents $\exp(\pm i a x)$, and transforming $i a x\to t$, you get an integral which is equivilent to the Incomplete gamma function.
Nathaniel Bubis
- 33,018