evaluate $\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})$

I need to prove $\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})$ for $0<Re(z)<1$. I tried $\cos t=\frac{e^{it}+e^{-it}}{2}$ and to integrate it along the contour from $\epsilon $ to R, then to $Ri$ via a quarter of a circle, then downward to $\epsilon i$, finally back to $\epsilon$ via a small quarter of a circle. However, I find that $e^{-it}t^{z-1} $ does not converge when $R\to \infty$, can anyone give a correction of my integration or offer another method to evaluate the integral?(would be better if using complex contours) thanks in advance.

Solution 1:

Use $$\cos \theta = \frac{1}{2} (e^{i\theta} + e^{-i\theta})$$ to split the integral into two parts, namely $$ \frac{1}{2} \int_0^\infty e^{it} t^{s-1} dt \quad\text{and}\quad \frac{1}{2} \int_0^\infty e^{-it} t^{s-1} dt.$$ Let $t^{s-1} = e^{(\log t)(s-1)}$ where the logarithm is the branch with the cut along the negative real axis.

To evaluate the first integral, let $t = iu$, so that $u$ runs along the negative imaginary axis, giving $$i \int_0^{-i\infty} e^{-u} e^{\log(iu)(s-1)} du.$$ Now we must have $\log(iu) = \log|u|$ because $iu$ is on the positive real axis. Furthermore $\log u = \log|u|-i\pi/2.$ Substituting, we have $\log(iu) = \log u +i\pi/2.$ This gives the integral $$i \int_0^{-i\infty} e^{-u} e^{(\log(u)+i\pi/2)(s-1)} du = e^{i\pi/2(s-1)} e^{i\pi/2} \int_0^{-i\infty} e^{-u} u^{s-1} du = e^{i\pi/2 s} \Gamma(s).$$

For the second integral, let $t = -iu$, so that $u$ runs along the positive imaginary axis, giving $$ -i \int_0^{+i\infty} e^{-u} e^{\log(-iu)(s-1)} du.$$ As before, $-iu$ is on the positive real axis so that $\log(-iu) = \log|u|.$ Furthermore, $\log u = \log|u|+i\pi/2.$ Substituting, we have $\log(-iu) = \log u -i\pi/2.$ This gives the integral $$-i \int_0^{+i\infty} e^{-u} e^{(\log(u)-i\pi/2)(s-1)} du = e^{-i\pi/2(s-1)} e^{-i\pi/2} \int_0^{+i\infty} e^{-u} u^{s-1} du = e^{-i\pi/2 s} \Gamma(s).$$ Put these two together to obtain $$\frac{1}{2}e^{i\pi/2 s} \Gamma(s) + \frac{1}{2} e^{-i\pi/2 s} \Gamma(s) = \cos(s \pi/2)\Gamma(s).$$ The computation of $\log(iu)$ and $\log(-iu)$ in terms of $\log u$ is necessary because in general we can't use $\log(zw) = \log z + \log w$ as with positive reals.

Remark. To complete the above argument you still need to show that those integrals along the imaginary axis in fact represent the Gamma function integral. Use a quarter-circle contour (first and fourth quadrant) to achieve this. I will do this myself later if no one solves it before I do and if it gets done sooner, even better.

Solution 2:

Substituting $e^{it}$ for $\cos(t)$ and taking the real part of the integral becomes complicated when we let $z$ be non-real, so I simply use $\cos(t)=\frac{e^{it}+e^{-it}}{2}$.

One must also justify the vanishing of the integral over the arc at infinity since that only vanishes when $\mathrm{Re}(z)\lt1$.

$$ \begin{align} \int_0^\infty\cos(t)\,t^{z-1}\,\mathrm{d}t &=\frac12\int_0^\infty e^{it}\,t^{z-1}\,\mathrm{d}t +\frac12\int_0^\infty e^{-it}\,t^{z-1}\,\mathrm{d}t\tag{1}\\ &=\frac12\int_0^{i\infty} e^{it}\,t^{z-1}\,\mathrm{d}t +\frac12\int_0^{-i\infty} e^{-it}\,t^{z-1}\,\mathrm{d}t\tag{2}\\ &=\frac12e^{\pi iz/2}\int_0^\infty e^{-t}\,t^{z-1}\,\mathrm{d}t +\frac12e^{-\pi iz/2}\int_0^\infty e^{-t}\,t^{z-1}\,\mathrm{d}t\tag{3}\\[9pt] &=\cos(\pi z/2)\Gamma(z)\tag{4} \end{align} $$ Justification:
$(1)$: $\cos(t)=\frac{e^{it}+e^{-it}}{2}$
$(2)$ left: integral over $[0,R]\cup Re^{i[0,\pi/2]}\cup[iR,0]$ is $0$
$\hphantom{(2)}$ right:integral over $[0,R]\cup Re^{i[0,-\pi/2]}\cup[-iR,0]$ is $0$
$(3)$: change variables
$(4)$: $\cos(t)=\frac{e^{it}+e^{-it}}{2}$

The integrals of $e^{it}\,t^{z-1}$ along $Re^{i[0,\pi/2]}$ and $e^{-it}\,t^{z-1}$ along $Re^{i[0,-\pi/2]}$ are bounded by $$ \begin{align} \int_0^{\pi/2}R^{\mathrm{Re}(z)}e^{\pi|\mathrm{Im}(z)|/2}e^{-R\sin(t)}\,\mathrm{d}t &\le\int_0^{\pi/2}R^{\mathrm{Re}(z)}e^{\pi|\mathrm{Im}(z)|/2}e^{-2Rt/\pi}\,\mathrm{d}t\\ &=\frac\pi2R^{\mathrm{Re}(z)-1}e^{\pi|\mathrm{Im}(z)|/2}(1-e^{-R})\\[6pt] &\to0 \end{align} $$ when $\mathrm{Re}(z)\lt1$.