Mellin transform-- complete Riemann function

[binbagsss]

Homework Statement

Show that $\int\limits^{\infty}_0 \sum\limits^{\infty}_{n=1}e^{-\pi n^{2} t} t^{s-1} dt = \sum\limits^{\infty}_{n=1}(\pi n^{2} )^{-s} \int\limits^{\infty}_{0} e^{-t} t^{s-1} dt$

Homework Equations

$\Phi (t)=\theta(it/2)=1+2\sum\limits^{\infty}_{n=1}e^{-\pi n^2 t}.$

With the transformation formula : $\Phi(1/t)=t^{1/2}\Phi(t)$

The Attempt at a Solution

I am unsure really where to start, a clue would be greatly appreciated.

The only thing I can really think of is transforming $t$ to get $t^{s}$ rather than $t^{1/2}$ on the RHS of (1), but I am a) unsure how to do this b) unsure whether it is the right thing to do

The other thing would be to expand out the exponential, writing it as two expressions multiplied together corresponding to the $-t$ in the exponent power and the $\pi n^{2}$ seperately, collecting the sum to infinity to get $e^{-t}$ and leaving the bit corresponding to "$e^{\pi n^{2}}"$ as a summation. However the $-s$ means the choice of my variable of this summation would need to be $s$, which doesn't really make sense?

I'm quite stuck.

Any help greatly appreciated,

Many thanks.

 

[eys_physics]

Homework Statement

Show that $\int\limits^{\infty}_0 \sum\limits^{\infty}_{n=1}e^{-\pi n^{2} t} t^{s-1} dt = \sum\limits^{\infty}_{n=1}(\pi n^{2} )^{-s} \int\limits^{\infty}_{0} e^{-t} t^{s-1} dt$

Homework Equations

$\Phi (t)=\theta(it/2)=1+2\sum\limits^{\infty}_{n=1}e^{-\pi n^2 t}.$

With the transformation formula : $\Phi(1/t)=t^{1/2}\Phi(t)$

The Attempt at a Solution

I am unsure really where to start, a clue would be greatly appreciated.

The only thing I can really think of is transforming $t$ to get $t^{s}$ rather than $t^{1/2}$ on the RHS of (1), but I am a) unsure how to do this b) unsure whether it is the right thing to do

The other thing would be to expand out the exponential, writing it as two expressions multiplied together corresponding to the $-t$ in the exponent power and the $\pi n^{2}$ seperately, collecting the sum to infinity to get $e^{-t}$ and leaving the bit corresponding to "$e^{\pi n^{2}}"$ as a summation. However the $-s$ means the choice of my variable of this summation would need to be $s$, which doesn't really make sense?

I'm quite stuck.

Any help greatly appreciated,

Many thanks.

Do a change of variables of the form $u=\pi n^2 t$.

 

[binbagsss]

Do a change of variables of the form $u=\pi n^2 t$.

cheers !