Let $\alpha \in (0,\infty) $. Calculate, if possible: $\int_1^\infty \sin(x) x^{-\alpha} \,dx$

Question

I have stumbled across an exercise in my math book but I'm unable to solve it:

Let $\alpha \in (0,\infty) $

$$\int_1^\infty \sin(x) x^{-\alpha} \,dx$$

There are are a few explanation online where $\alpha = 1$, but never the general form for $\alpha \in (0,\infty) $.

Answer 1

There is a closed form if you required that $\alpha \in ]0,1[$ and the integral beeing from $0$ to $+\infty$.

Partial answer :

Let $g_{\alpha} : \mathbb{R} \times ]0,+\infty[ \hspace{0.1cm} \longmapsto \mathbb{C}$, $g_{\alpha}(x,y) = y^{\alpha -1}e^{-xy}e^{ix}$, and $\forall b > 0$ we define $S_{b} :=]0,b] \times ]0,+\infty[$.

Using Tonelli since $|g_{\alpha}(x,y)| = y^{\alpha-1}e^{-xy}$ we have that on $S_{b}$ :

$$\int_{y=0}^{y=\infty}y^{\alpha-1}\int_{0}^{b}e^{-xy}dxdy = \int_{0}^{\infty}y^{\alpha-1}\frac{1-e^{-yb}}{y}dy$$

We have $\lim\limits_{y \to 0^{+}} \frac{(1-e^{-yb})}{y} = b$, which means that the integrand is asymptotic to $y^{\alpha-1}b = \frac{b}{y^{1-\alpha}}$, integrabile in a neighbourhood of $0$ since $1-\alpha < 1$, being $\alpha > 0$.

When $y \to +\infty$ instead the integrand is asymptotic to $\frac{y^{\alpha -1}}{y} = \frac{1}{y^{2-\alpha}}$, integral around infinity since $2-\alpha > 1$ because $\alpha < 1$.

Thanks to Fubini we have that the order of integration in $I_{\alpha}(b) := \int_{S_{b}} g_{\alpha}(x,y)dxdy$ can be interchanged.

Additionally let's see that :

(i) $\lim\limits_{b \to \infty} \int_{0}^{+ \infty} y^{\alpha-1} \frac{e^{-yb}}{i-y} = 0$. To notice this we see that $|y^{\alpha-1} \frac{e^{-yb}}{i-y}| = y^{\alpha-1}\frac{e^{-yb}}{(1+y^{2})^{\frac{1}{2}}}$, Integrable on $[0,+\infty[$, and by monotone convergence since the integral of the absolute value when $b\to +\infty$ tends decreasing to $0$, we have that the integral itself goes to $0$.

(ii) Let $t = xy$ we have $$\int_{0}^{\infty}y^{\alpha-1}e^{-xy}dy = \frac{1}{x^{\alpha}}\int_{0}^{+\infty}t^{\alpha-1}e^{-t}dt = \frac{\Gamma(\alpha)}{x^{\alpha}}$$

Now, beeing

$\int_{0}^{b} e^{-xy}e^{ix}dx = \int_{0}^{b}e^{(i-y)x}dx = \frac{e^{(i-y)b}-1}{i-y}$

We get that

$I_{\alpha}(b) = \int_{0}^{\infty} y^{\alpha-1} \frac{e^{(i-y)b}-1}{i-y} dy = e^{ib} \int_{0}^{\infty} y^{\alpha -1} \frac{e^{-yb}}{i-y}dy - \int_{0}^{\infty} \frac{y^{\alpha -1}}{i-y}dy$

Letting $b \to +\infty$, as we saw $\lim\limits_{b \to \infty} \int_{0}^{+ \infty} y^{\alpha-1} \frac{e^{-yb}}{i-y} = 0$ and because $e^{ib}$ keeps bounded we have

$$\lim\limits_{b \to \infty} I_{\alpha}(b) = \int_{0}^{\infty} \frac{y^{\alpha -1}}{y-i}dy = \int_{0}^{\infty} \frac{y^{\alpha}}{1+y^{2}}dy +i\int_{0}^{\infty} \frac{y^{\alpha-1}}{1+y^{2}}dy$$

From Fubini instead we have that $I_{\alpha}(b) = \int_{x=0}^{x=b}e^{ix}\int_{0}^{+\infty} y^{\alpha-1}e^{-xy}dydx$ and from the previous observation in terms of gamma function we have that $I_{\alpha}(b) = \Gamma(\alpha) \int_{0}^{b} \frac{e^{ix}}{x^{\alpha}}dx$

In other words $$\lim\limits_{b \to +\infty} I_{\alpha}(b) = \Gamma(\alpha)(\int_{0}^{+\infty}\frac{cosx}{x^{\alpha}}dx+i\int_{0}^{+\infty}\frac{sinx}{x^{\alpha}})$$

If in the integral $\int_{0}^{+\infty} \frac{y^{\alpha-1}}{1+y^{2}}$ we set $y^{2} = t$ we get $\frac{1}{2} \int_{0}^{\infty}\frac{t^{\frac{\alpha}{2}}-1}{1+t}dt = \frac{1}{2} B(\frac{\alpha}{2},1-\frac{\alpha}{2}) = \frac{\pi}{2sin(\frac{\pi \alpha}{2})}$

(Where $B(p,q) = \int_{0}^{+\infty} \frac{t^{p-1}}{(1+t)^{p+q}}$)

Last but not least we get in the end $$\int_{0}^{+\infty}\frac{sinx}{x^{\alpha}} dx = \frac{\pi}{2\Gamma(\alpha)sin(\frac{\pi \alpha}{2})}$$

Answer 2

According to CAS the integral evaluates to

$$\int_1^\infty \frac{\sin(x)}{x^a} \,dx=\frac{_1F_2\left(1-\frac{a}{2};\frac{3}{2},2-\frac{a}{2};-\frac{1}{4}\right)}{a-2}+\cos\left(\frac{\pi a}{2}\right) \Gamma (1-a),$$ where $_1F_2(\dots)$ and $\Gamma(x)$ are the hypergeometric and gamma functions, respectively.