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

calculuscontour-integrationdefinite integralsintegrationtrigonometric-integrals

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?

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