Show that $\int_{0}^{\infty}\frac{t^{\tau - 1}}{1+t}= \frac{\pi}{\sin(\pi \tau)}$, where $0<Re(\tau)<1$

Question

I need to show that Mellin Transform of the function $\frac{1}{1+t}$ is $\frac{\pi}{\sin(\pi \tau)}$. So, by definition $(Mf)(\tau)=\int_{0}^{\infty}f(t)t^{\tau -1}dt=\int_{0}^{\infty}\frac{t^{\tau - 1}}{1+t}dt$. In my attemp, i choose $u=t^{\tau -1}$ and $dv=\frac{dt}{1+t}$ hence $du=(\tau -1)t^{\tau - 2}$ and $v=\log(1+t) $ and $\int_{0}^{\infty}\frac{t^{\tau - 1}}{1+t}dt=(t^{\tau -1}\log(1+t))|_{0}^{\infty}-(\tau-1)\int_{0}^{\infty}t^{\tau -2}\log(t+1)dt$. Next suppose that $\lim_{t\to \infty}\frac{\log(1+t)}{t^{1-\tau}}=0$, because $t^{1-\tau}$ is faster than $\log(1+t)$ (Suppose because $Re(\tau)<1$), i mean, i want to use that $t^{1-\tau}=e^{\log(t^{1-\tau})}=e^{(1-\tau)\log t}=e^{(1-Re(\tau))\log t}e^{i\log t}$ but $|e^{i\log t}|=1$...After that my new integral is $(Mf)(\tau)=-(\tau -1)\int_{0}^{\infty}t^{\tau -2}\log(t+1)dt$ and again choose $u=t^{\tau -2}$ and $dv=\log(1+t)dt$ then $du=(\tau -2)t^{\tau -3}$ and $v=(1+t)(\log(1+t)-1)$. Imply $(Mf)(\tau)=-(\tau -1)[t^{\tau -2}(1+t)(\log(1+t)-1)|_{0}^{\infty}-(\tau -2)\int_{0}^{\infty}t^{\tau -3}(1+t)(\log(1+t)-1)]$. Need i continous like this or need other way like contours when complex integral use Cauchy?..well i note that $(\tau -1)(\tau -2)...=\Gamma(\tau -1)$ and by properties of Gamma function $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin \pi z}$ but how to arrive there? thanks

Answer 1

Hint: the simplest way to evaluate the integral is by contour integration. Use the cut along the positive real semi-axis and choose the contour running from $0$ to $+\infty $ along the upper bank of the cut, then along the full circle centered at $z=0$ and finally from $+\infty$ back to $0$ along the lower bank of the cut.