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

Solution 1:

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

Solution 2:

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: