\(\int_{0}^{+\infty}{x^{-a}\sin{x}\,dx}\)

[Grigorios Kostakos]

For \(a\in(0,2)\) prove that \[\displaystyle\int_{0}^{+\infty}{\frac{\sin{x}}{x^a}\,dx}=\Gamma(1-a)\,\cos\big(\tfrac{\pi a}{2}\big)\,,\]

where \(\Gamma(z)\) is Gamma function.





P.S. I have a rather complicated proof for the case \(0<a<1\). So i seek, if it's possible, a more elegant proof.

[Tolaso J Kos]

For \(a\in(0,2)\) prove that \[\displaystyle\int_{0}^{+\infty}{\frac{\sin{x}}{x^a}\,dx}=\Gamma(1-a)\,\cos\big(\tfrac{\pi a}{2}\big)\,,\]

where \(\Gamma(z)\) is Gamma function.

Hello Grigoris,

here is a solution. One can actually begin by the Mellin Transformation of the \( \sin \) function. So,

$$\begin{aligned}
\int_0^\infty x^{a-1}\sin(x)\,\mathrm{d}x
&=\frac1{2i}\left(\int_0^\infty x^{a-1}e^{ix}\,\mathrm{d}x-\int_0^\infty x^{a-1}e^{-ix}\,\mathrm{d}x\right)\\
&=\frac1{2i}\left(e^{ia\pi/2}\int_0^\infty x^{a-1}e^{-x}\,\mathrm{d}x-e^{-ia\pi/2}\int_0^\infty x^{a-1}e^{-x}\,\mathrm{d}x\right)\\
&=\sin\left(\frac{a\pi}{2}\right)\Gamma(a)
\end{aligned}$$

Setting \( a \mapsto 1-a \) we get that:

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

which is the desired formula.

P.S 1: The formula does not hold for \( a= 1\) since the LHS is \( \pi/2 \) and the RHS is not defined. Neverthess, the formula does hold true for all the other numbers. For example for \( a=3/2 \) we have that:

$$\int_{0}^{\infty}\frac{\sin x}{x^{3/2}}\, {\rm d}x = \Gamma \left ( 1- \frac{3}{2} \right )\cos \frac{3\pi}{4}= - \frac{\sqrt{2}}{2}\cdot \Gamma \left ( - \frac{1}{2} \right ) = \sqrt{2\pi}$$

which is true. Of course \( \Gamma (-z) \) is evaluated through the formula:

$$\Gamma(x)\Gamma(-x)= - \frac{\pi}{x \sin \pi x}$$

So the formula does hold true when \( a \in (0, 1) \cup (1, 2) \).

P.S 2: The changes of variables \( x\mapsto ix \) and \( x\mapsto-ix \) used in the second equation above, are justified because
$$\oint_{\gamma_k} z^{a-1}e^{-z}\,\mathrm{d}z=0 $$

where \( \gamma_1=[0,R]\cup Re^{i\frac\pi2[0,1]}\cup iR[1,0] \) and \( \gamma_2=[0,R]\cup Re^{-i\frac\pi2[0,1]}\cup-iR[1,0] \) contain no singularities.