Solving Gamma Function Integral: e^(4u)*e^(-e^u)du

[splitendz]

Hi. I'm having some trouble solving the following gamma function:

Evaluate the integral e^(4u) * e^(-e^u)du. The upper limit is inifinity and the lower limit is 0.

I'm letting x = e^(u) or u = 1 in the hope to have the function looking similar to the gamma function. But I'm having no luck as du/dx will be equal to zero in this case.

Help please! :)

 

[James R]

You want

$$\int \limits_0^\infty e^{4u}e^{-e^u} du$$

Put $x=e^u$, so that $dx=e^u du$ and you get:

$$\int \limits_1^\infty x^3 e^{-x} dx$$

Can you go from there?

 

[PhysicsinCalifornia]

You want

$$\int \limits_0^\infty e^{4u}e^{-e^u} du$$

Put $x=e^u$, so that $dx=e^u du$ and you get:

$$\int \limits_1^\infty x^3 e^{-x} dx$$

Can you go from there?

How do you write those sophisticated symbols?

I want to be able to write it, but don't know how

(Sorry its not helping your problem dude)

 

[splitendz]

Thanks James. I'm right to continue now but shouldn't it be x^4 not x^3?

 

[Moo Of Doom]

Actually, dx=e^u du, so one of the e^u's is in the dx, leaving only e^3u, or x^3.

 

[splitendz]

Of course. Thanks for your help guys :) :)

 

[dextercioby]

I think part integrating will do it.Three times,i guess.

Daniel.

 

[HallsofIvy]

How do you write those sophisticated symbols?

I want to be able to write it, but don't know how

(Sorry its not helping your problem dude)

If you go to the "general physics" forum you will find a "sticky" on Latex formatting.