Mellin Transform of $e^{-2 \pi t}$

Question

I need to show that if $f(\tau)=e^{-2 \pi \tau}$ then:

$$\{\mathcal{M}\,f\}(s)=(2 \pi)^{-s} \Gamma(s)$$

where :

$$\Gamma(s)=\int_{0}^{+\infty} e^{-t}t^{s}\frac{dt}{t}$$

and

$$\{\mathcal{M}\,f\}(s)=\int_{0}^{+\infty} f(t)\,t^{s}\frac{dt}{t}$$

But I do not know which change of variables is the correct thank you :)

Answer 1

The mellin transform is, as far as I know, with $\;f(t)\;$ inside the integral, not of $\;f(it)\;$:

$$\{f\}(s):=\int\limits_0^\infty e^{2\pi t}t^{s-1}dt$$

Change of variable:

$$u=-2\pi t\implies dt=-\frac{du}{2\pi }$$

$$=\int\limits_0^\infty e^{-u}\left(-\frac{u}{2\pi }\right)^{s-1}\left(-\frac{du}{2\pi }\right)=(2\pi)^{-s} \int\limits_0^\infty e^{-u}u^{s-1}du$$

and we get what we want.

Answer 2

$$\int_{0}^{+\infty}e^{-2\pi x}x^{s-1}\,dx = \frac{1}{(2\pi)^{s}}\int_{0}^{+\infty}x^{s-1}e^{-x}\,dx = \frac{\Gamma(s)}{(2\pi)^s}.$$