Interesting integrals, which I think involve the gamma function

1. Mar 4, 2012

MeMoses

1. The problem statement, all variables and given/known data
Evaluate the intergrals:
a) integral of 3^(-4*z^2) dz from 0 to infinity
b) integral of dx/(sqrt(-ln(x))) from 0 to 1
c) integral of x^m * e^(-a*x^n) dx from 0 to infinity

2. Relevant equations

gamma(n) = integral of e^(-w) * w^(n-1) dw from 0 to infinity

3. The attempt at a solution
I'm assuming I use:
gamma(n) = integral of e^(-w) * w^(n-1) dw from 0 to infinity
as it was used in the rest of the problems in this set, however I have no idea where to begin on (a) and (b). For (c) I've been trying to get it to match up with the gamma function above, but the a variable is giving me some difficulty. I should be able to get (c) eventually, but if you could please help with (a) and (b) that would be great as well with anything that simplifies (c)

2. Mar 4, 2012

Dick

For a) change 3^(-4*z^2) to e^(log(3)*(-4)*z^2), now try the substitution w=log(3)*4*z^2. For b) the obvious thing to try is w=(-ln(x)), do you see why?

3. Mar 4, 2012

MeMoses

for a) once w=4z^2*ln(3), dw=8ln(3)z, so then i get
1/(8ln(3)) integral of e^(-w)*z^-1 dw. So do I solve w^(n-1) = z^-1 for n and take the gamma function of n? That just seems wrong/overly comlicated for me.
As for b) I'm not sure what there is to see, and when do the limits change to match those on the integral on the gamma function? Also, I'm currently stuck on c) as well

4. Mar 4, 2012

SammyS

Staff Emeritus
For (a):

Let $u=2(\sqrt{ln(3)})z\,.$

Look at http://en.wikipedia.org/wiki/Gaussian_integral if you don't know the result for the Gaussian Integral.

5. Mar 4, 2012

SammyS

Staff Emeritus
For (b):

Do integration by parts.
Choose u & dv in a somewhat backwards way.

This is what you want for v : $\displaystyle v =\sqrt{-ln(x)}\,.$

Then find dv and find what u must be.

Why do you want $\displaystyle v =\sqrt{-ln(x)}\,?$
Look at the graph of $f(x)=\sqrt{-ln(x)}\,.$

The integral $\displaystyle \int_0^1 f(x)\,dx$ is the same as $\displaystyle \int_0^\infty f^{-1}(y)\,dy$

$\displaystyle f^{-1}(y)=\underline{\ ?\ }$​

6. Mar 4, 2012

Dick

Hey SammyS, you can do all of these by substituting and putting into the form of a gamma function. Your alternative works, but the gamma route doesn't involve any integration by parts.

7. Mar 4, 2012

SammyS

Staff Emeritus
Thanks !