Proving the asymptotic of $\int_0^\infty\frac{\sin x}{x^{a+1}}\,dx$ for $a \to 0^+$ is $\pi/2$

Question

I am working on finding the asymptotic of $\int_0^\infty\frac{\cos x}{x^a} \, dx$ for $a\to0^+$. And I have reduced the problem on finding the asymptotic of $a\int_0^\infty\frac{\sin x}{x^{a+1}} \,d x$. The result is known to be $a\pi/2$ if $a\to0^+$ and there is my proof. $$a\int_0^\infty\frac{\sin x}{x^{a+1}} \,dx = a\int_0^{\pi/2}\frac{\sin x}{x^{a+1}} \, dx + a\int_{\pi/2}^\infty\frac{\sin x}{x^{a+1}} \, dx$$ On the other hand $a\int_0^{\pi/2}\frac{\sin x}{x^{a+1}}\,dx$ behaves as $a\pi/2$ and $a\int_{\pi/2}^\infty\frac{\sin x}{x^{a+1}} \, dx = O(1)$. However the problem is that $a\pi/2$ is way smaller that $1$. Any suggestions would be appreciated.

Answer 1

Suppose $0<a<1$. Since $x^{a}\sin x\xrightarrow{x\rightarrow0}0$, integration by parts gives

$$ \int^M_0x^{-a}\cos x\,dx = M^{-a}\sin M +a\int ^M_0x^{-a-1}\sin x\,dx $$

Sine $x^{-a-1}(1-\cos x)\xrightarrow{x\rightarrow0}0$, following Daniel Fischer's suggestion (integration by parts using $(1-\cos x)'=\sin x$) gives

$$ \int^M_0x^{-a-1}\sin x\,dx = M^{-a-1}(1-\cos M) + (a+1)\int^M_0x^{-a-2}(1-\cos x)\,dx $$

Putting things together and letting $M\rightarrow \infty$ gives

$$ \int^M_0x^{-a}\cos x\,dx = a(a+1)\int^\infty_0\frac{1-\cos x}{x^{a+2}}\,dx $$

The integral to the right, as a function of $a$, can be handled by dominated convergence arguments since $0\leq 1-\cos x\leq 2$, $\frac{1-\cos x}{x^2}$ is bounded in $(0,1]$, and $x^{-1-a}$ converges to $x^2$ in $L_2(1,\infty)$. Thus $\int^\infty_0 \frac{1-\cos x}{x^{2+a}}\xrightarrow{a\rightarrow0} \int^\infty_0\frac{1-\cos x}{x^2}\,dx$.

It follows that $\lim_{a\rightarrow0}\int^\infty_0 x^{-a}\cos x\,dx =0$.