How to prove the following? $$\int_0^\infty \frac{1-\cos(at)}{t^{1+\alpha}} \mathrm dt =\frac{\pi}{2 \Gamma(\alpha+1) \sin(\alpha \pi /2)} |a|^\alpha $$
I have tried to do it with some simple methods such as integration by parts or something else and transformed it into calculate $ \int_0^{2\pi}\sin^a\theta\cos^{1-a}\theta d\theta $ .
Still I can't solve it. Well, I know I'm not gonna solve it through such simple ways I've tried, but I'm indeed not skillful in calculus.
Using $$ L(x^\alpha, t)=\frac{\Gamma(\alpha+1)}{t^{\alpha+1}}, L(\cos(at),x)=\frac{x}{x^2+a^2}$$ and $$\int_0^\infty \frac{x^{\alpha-1}}{x^2+a^2}\mathrm dx=\frac{|a|^\alpha\pi}{2\sin(\frac{\alpha\pi}{2})} $$ one has \begin{eqnarray} &&\int_0^\infty \frac{1-\cos(at)}{t^{1+\alpha}} \mathrm dt\\ &=& \frac1{\Gamma(\alpha+1)} \int_0^\infty L(x^\alpha,t) (1-\cos(at))\mathrm dt\\ &=& \frac1{\Gamma(\alpha+1)}\int_0^\infty x^\alpha L(1-\cos(at),x)\mathrm dx\\ &=& \frac1{\Gamma(\alpha+1)}\int_0^\infty x^\alpha(\frac{1}{x}-\frac{x}{x^2+a^2})\mathrm dx\\ &=& \frac1{\Gamma(\alpha+1)}\int_0^\infty \frac{x^{\alpha-1}}{x^2+a^2}\mathrm dx\\ &=& \frac1{\Gamma(\alpha+1)}\frac{|a|^\alpha\pi}{2\sin(\frac{\alpha\pi}{2})}\\ &=&\frac{\pi}{2 \Gamma(\alpha+1) \sin(\frac{\alpha \pi}{ 2})} |a|^\alpha. \end{eqnarray} Here $L(f(x),t)$ is the Laplace transform of $f(x)$ or $$ L(f(x),t)=\int_0^\infty f(x)e^{-tx}dx. $$ One property of Laplace transform is $$ \int_0^\infty L(f(x),t)g(t)dt=\int_0^\infty f(x)L(g(t),x)dx. $$