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Old Aug 23rd 2018, 07:38 PM   #11
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Originally Posted by topsquark View Post
$\displaystyle = \sum_{n = 1}^{\infty} \int x^3 e^{-nx} ~ dx$
Okay, from here we put back the limits on the integration. Do integration by parts three times and you get:
$\displaystyle \int_0^{\infty} x^3 e^{-nx} ~ dx = \frac{6}{n^4}$

Thus we want
$\displaystyle \sum _{n = 1}^{\infty} \int_0^{\infty} x^3 e^{-nx} ~ dx = \sum _{n = 1}^{\infty} \frac{6}{n^4} $

If you know something about special functions you will note that this summation is equal to the Riemann zeta function.
$\displaystyle \zeta (s) = \sum _{n = 1}^{\infty} \frac{1}{n^s}$

Some of these take on special values. For example, your sum
$\displaystyle \zeta (4) = \sum _{n = 1}^{\infty} \frac{1}{n^4} = \frac{\pi ^4}{90}$

Thus
$\displaystyle \sum _{n = 1}^{\infty} \int_0^{\infty} x^3 e^{-nx} ~ dx = \frac{6}{n^4} = \frac{6 \pi ^4}{90}$

I'll let you pull what remains together, as well as putting those pesky constants back in.

If you need further discussion, just let me know.

-Dan
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Old Aug 24th 2018, 11:43 AM   #12
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Hello Dan.

Thanks.

I did a U sub to get the Integral in the form you suggested then followed your instructions and it works!!

I do have one final question about using L'Hopital rule...

After I got my indefinite Integral from Integration by Parts, I used L'Hopital rule to calculate the definite integral with the boundaries.

Calculating the limit at 0 was easy, and gave me the result you quoted, but I wasn't sure about the process for the boundary value at infinity.

I had a denominator of the form ne^nx which I wanted to differentiate and substitute in infinity. I concluded that if I kept differentiating this term the x in the index term to e would never disappear, and hence it would always equal infinity at the boundary condition so the terms at the Infinity boundary would tend to 0. Is that the right way to approach this?

(I am doing a Masters Astrophysics and Astronomy course in September but I come from a Geology background and don't have the Maths experience so I'm struggling a little. Trying to play catch up. So thanks very much for helping me. I really appreciate it.)
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Old Aug 24th 2018, 02:10 PM   #13
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Originally Posted by Mazin View Post
Hello Dan.

Thanks.

I did a U sub to get the Integral in the form you suggested then followed your instructions and it works!!

I do have one final question about using L'Hopital rule...

After I got my indefinite Integral from Integration by Parts, I used L'Hopital rule to calculate the definite integral with the boundaries.

Calculating the limit at 0 was easy, and gave me the result you quoted, but I wasn't sure about the process for the boundary value at infinity.

I had a denominator of the form ne^nx which I wanted to differentiate and substitute in infinity. I concluded that if I kept differentiating this term the x in the index term to e would never disappear, and hence it would always equal infinity at the boundary condition so the terms at the Infinity boundary would tend to 0. Is that the right way to approach this?

(I am doing a Masters Astrophysics and Astronomy course in September but I come from a Geology background and don't have the Maths experience so I'm struggling a little. Trying to play catch up. So thanks very much for helping me. I really appreciate it.)
How do you do the tall vertical line in LaTeX. (See the short version below next to the pq.)

By the numbers, then.
$\displaystyle \int _0^{\infty} x^3 e^{-nx} ~ dx$

We have the form $\displaystyle \int _a^b p ~dq = pq \lvert _a^b - \int _a ^b q~dp$

So pick $\displaystyle p = x^3$ and $\displaystyle dq = e^{-nx}~dx$
$\displaystyle \lim _{a \to \infty} \int _0^a x^3 e^{-nx} ~ dx = \lim_{a \to \infty} \left ( x^3 ~ \frac{-1}{n} e^{-nx} \lvert _0^a - \int _0^a 3x^2 \frac{-1}{n} e^{-nx} ~ dx \right )$

The first term on the RHS (ignoring the -1/n):
$\displaystyle \lim _{a \to \infty} x^3 e^{-nx}\lvert _0^a = \lim_{a \to \infty} a^3 ~ e^{-na} - (0^3) e^{0} = \lim_{a \to \infty} a^3 ~ e^{-na}$

To calculate the limit, your method is good and L'Hopital's rule is good. Here's another way, which can be made rigorous but I call this "Physics Math."

First
$\displaystyle e^{na} = 1 + na + \frac{n^2}{2} a^2 + \frac{n^3}{6} a^3 + \frac{n^4}{24} ~ a^4 \text{ + .... }$
(This line is a little sketchy as the sum doesn't converge for large a, but like I said, Physics Math. The argument makes sense otherwise.)

So
$\displaystyle \lim_{a \to \infty} a^3 e^{-na} = \lim_{a \to \infty} \frac{a^3}{1 + na + \frac{n^2}{2} a^2 + \frac{n^3}{6} a^3 + \frac{n^4}{24} ~ a^4 \text{ + .... }}$

$\displaystyle = \lim_{a \to \infty} \frac{1}{\frac{1}{a^3} + \frac{n}{a^2} + \frac{1}{na} + \frac{n^3}{6} + \frac{n^4}{24} a \text{ + ... }} = 0$

thus $\displaystyle \lim_{a \to \infty} x^3 e^{-nx} \lvert _0^a = 0$

and thus
$\displaystyle \lim _{a \to \infty} \int _0^a x^3 e^{-nx} ~ dx = \frac{3}{n} \lim_{a \to \infty} \int _0^a x^2 e^{-nx} ~ dx $

The other two integration by parts and limits are done by the same methods.

-Dan
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Last edited by topsquark; Aug 24th 2018 at 02:49 PM.
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