Definite integral of a power times a cosine

Question

The following integral appears in Gradshteyn & Rhyzik, page 421 3.769 $$\int_0^{\infty } x^{\mu -1} \cos (a x) \, dx=\frac{\Gamma (\mu ) \cos \left(\frac{\mu \pi }{2}\right)}{a^{\mu }}$$ (a>0, $0<\Re(\mu )<1$)

I am looking for a derivation

Answer 1

As pointed out above,the integral in question is obtained trivially from Ramanujan's Master Theorem. Another approach is to use contour integration with f(z)=$z^{\mu -1} e^{i a z}$ where the contour consists of the two circular arcs $C_{\epsilon }$ and $C_R$ in the first quadrant along with the segements from ($\epsilon$,0) to (R,0) and (0,$\epsilon$) to (0,R). The intergral along $C_{\epsilon }$ approaches zero as $\epsilon \to 0$ and the integral along $C_R$ approaches zero as $R\to \infty$. The intgral along the segment of the real axis becomes $\int_0^{\infty } x^{\mu -1} e^{i x} \, dx$ while he intgral along the segment of the imaginary axis assumes the form $e^{\frac{i \pi \mu }{2}} \Gamma (\mu )$. Taking the real parts of both expressions yields the required integral.

Answer 2

Well actually the mentioned theorem holds true for all $\Re(\mu)>0$. Well there are a couple of methods using complex Analysis but there is also an approach using simple real analysis. Start with the integral definition of gamma function, $$\Gamma(s)=\int_{0}^{\infty} t^{s-1}e^{-t}dt, \Re(s)>0$$ Substitute, $t=pu^{n} \implies dt=pnu^{n-1}du$ $$\Gamma(s)=np^{s} \int_{0}^{\infty} u^{ns-1} e^{-pu^{n}}du$$ Divide by $np^{s}$ on both sides of equation, $$\frac{\Gamma(s)}{np^{s}}=\int_{0}^{\infty}u^{ns-1}e^{-pu^{n}} du \rightarrow (1)$$ Now here's the trick. Let $p$ be a complex number. It sounds weird , as we did everything under assumption that $p$ was real. But you can actually verify it to be true by taking some examples. Now let, $p=a+ib=|p|e^{i\alpha}$ Similar to $(1)$ consider the complex conjugate of $p$. And rewrite the integral, $$\frac{\Gamma(s)}{n(\bar{p})^{s}} \int_{0}^{\infty} u^{ns-1}e^{-\bar{p}u^{n}}du \rightarrow (2)$$ Add $(1),(2)$ and use $p=a+ib=|p|e^{i\alpha}$. $$\frac{\Gamma(s)e^{-is\alpha}}{n|p|^{s}}+\frac{\Gamma(s)e^{is\alpha}}{n|p|^{s}}=\int_{0}^{\infty} u^{ns-1}e^{-au^{n}}(e^{ibu^{n}}+e^{-ibu^{n}})du$$ By using the Euler's formula, $$\frac{\Gamma(s)\cos(s\alpha)}{n|p|^{s}} =\int_{0}^{\infty}u^{ns-1}e^{-at^{n}}\cos(bu^{n})du$$ This result is powerful than expected. You can prove stuff like Dirichlet integral, Fresnel integral and others which require complex Analysis using this simple integral. $$\frac{\Gamma(s)\cos(s\alpha)}{n|p|^{s}}=\int_{0}^{\infty}u^{ns-1}e^{-au^{n}}\cos(bu^{n})du$$ Adjust the values of $p,n$ as required. In this case, set $a=0, p=ib, n=1$. Since real part is $0$, argument equals $\frac{\pi}{2}$. Putting all the values In the integral above $$\frac{\Gamma(s)\cos(s\frac{\pi}{2})}{b^{s})} =\int_{0}^{\infty}u^{s-1}\cos(bu) du$$ Since the integral definition of gamma function we considered converges for all $\Re(s) \gt 1$, we have that $$\frac{\Gamma(s)\cos(s\frac{\pi}{2})}{b^{s}}=\int_{0}^{\infty}u^{s-1}\cos(bu) du$$ For all $\Re(s)>0$.

The integral I derived is called as Euler integral. I found it here