Integral $\int_0^\infty x^{-\alpha} \sin(x)dx.$

Question

I am trying to understand how to integrate $$\int_0^\infty x^{-\alpha} \sin(x)dx.$$ Mathematica indicates that the integral exists for $0< \alpha < 2$ and gives $$ \dots = \cos(\alpha \pi/2)\Gamma(1-\alpha)$$ where $\Gamma$ is the gamma function. I have tried looking the integral up in Gradshteyn's book but I could not find it. Is there some obvious trick?

Answer 1

Consider the Mellin transform of the sine function:

$$\left\{\mathscr{M} \sin\right\}(s) = \int_{0}^{\infty} x^{s-1} \sin(x) dx $$

We can expand $\sin x$ using the Euler's formula and the de Moivre theorem:

$$ \sin(w) = \frac{e^{ix}-e^{-ix}}{2i} = \sum_{n=0}^{\infty} \frac{(ix)^n}{2in!} -\sum_{n=0}^{\infty} \frac{(-ix)^n}{2in!}=\sum_{n=0}^{\infty} \frac{(-x)^n\left[(-i)^n-(i)^n\right]}{2in!} $$

By de Moivre's theorem

$$(-i)^{n} = \cos\left(\frac{n\pi}{2}\right)-i\sin\left(\frac{n\pi}{2}\right)$$

$$i^{n} = \cos\left(\frac{n\pi}{2}\right)+i\sin\left(\frac{n\pi}{2}\right) $$

Therefore

$$\sin x = \sum_{n=0}^{\infty}\frac{(-x)^n[-\sin\left(\frac{n\pi}{2}\right)]}{n!}dw $$

Applying the Ramanujan's master theorem and using the fact that sine is odd

$$\left\{\mathscr{M} \sin\right\}(s) = \int_{0}^{\infty} x^{s-1} \sin(x) dx = \Gamma(s)\sin\left(\frac{s\pi}{2}\right) $$

If $\displaystyle s = 1-a$ we have

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

Note that $\Gamma(1-a)$ is defined for $a\neq 1$.

Therefore

$$\boxed{\int_{0}^{\infty} x^{-a} \sin(x) dx = \Gamma(1-a)\cos\left(\frac{a\pi}{2}\right) \quad a\in(0,1)\cup(1,2)} $$

For the case $a=1$ the integral is the usual Dirichlet integral