I I need integrals Int[0,infty]t^(a-1) e^(-t) F(z, e^(-t))dt

[benorin]

TL;DR Summary

I need your integrals which are interesting, of the form Int[0,infty]t^(a-1) e^(-t) F(z, e^(-t))dt and that possess known analytical solutions please?

I’m doing some brainstorming for a note I’m writing, I would appreciate it if anybody knows interesting integrals of the form

$$\int_{t=0}^\infty t^{\alpha - 1} e^{-t} F(z, e^{-t})\, dt=G( z, \alpha )$$

where $z$ and $\alpha$ are complex parameters and the solution $G(z, \alpha )$ is known? Even if they seem kinda simple ones, example:

$$\int_{t=0}^\infty t^{\alpha - 1} e^{-t} \cos (t)\, dt= 2^{\tfrac{-\alpha}{2}}\cos \left(\tfrac{\pi}{4}\alpha\right) \Gamma (\alpha )$$

@fresh_42 you always come up with those juicy integrals - if you’re able maybe come post one? Thanks.

If I end up using any of the integrals you guys come up with I will cite this thread and your handle or real name if you prefer?

-Ben

 

[anuttarasammyak]

It seems Mellin transform of $e^{-t} F$.　　
F=e^t M^{-1}\{G\}
where $M^{-1}$ is inverse Mellin transform.

 

[fresh_42]

Here is an example where the Gamma function is used:
https://www.physicsforums.com/threads/micromass-big-integral-challenge.867904/#post-5451293
and here is the book
https://www.physicsforums.com/threads/micromass-big-integral-challenge.867904/#post-5449982
those problems were from.

 

[benorin]

It seems Mellin transform of $e^{-t} F$.　　
F=e^t M^{-1}\{G\}
where $M^{-1}$ is inverse Mellin transform.

This is a good observation, thank you, but I don't understand how the compound function is accounted for in this? I mean the $e^{-t}$ inside $F( z, e^{-t})$ inside the integral transform in the first post? I am uncertain how to work with this feature, I suppose I could simply define another function $H( z,t) := F( z, e^{-t} )$ and use that somehow. How would I use a table of transforms for this kind of function?

 

[anuttarasammyak]

I am not sure whether I got you but
e^tM^{-1}\{G{(z,\alpha} )\}:= H(z,t)
H(z,t)=H(z,-\log(e^{-t}))=F(z,e^{-t}).
by substituting t with -log(e^-t) in H.