Integration question

• Jun 25th 2013, 10:10 PM
downthesun01
Integration question
Can someone show me how

$\displaystyle \int_{0}^{\infty}t^{k}e^{-at}dt=\frac{k!}{a^{k+1}}\text{ where }k\geq 0 \text{ is an integer and }a>0$

I understand that you repeatedly integrate by parts, but how does everything reduce to $\displaystyle \frac{k!}{a^{k+1}}$?

Thanks
• Jun 26th 2013, 01:05 AM
JJacquelin
Re: Integration question
Quote:

Originally Posted by downthesun01
Can someone show me how

$\displaystyle \int_{0}^{\infty}t^{k}e^{-at}dt=\frac{k!}{a^{k+1}}\text{ where }k\geq 0 \text{ is an integer and }a>0$

I understand that you repeatedly integrate by parts, but how does everything reduce to $\displaystyle \frac{k!}{a^{k+1}}$?

Thanks

Did you repeatedly integrate it by parts and see what happend ? Show us what you have done.
• Jun 26th 2013, 01:55 AM
Prove It
Re: Integration question
Quote:

Originally Posted by downthesun01
Can someone show me how

$\displaystyle \int_{0}^{\infty}t^{k}e^{-at}dt=\frac{k!}{a^{k+1}}\text{ where }k\geq 0 \text{ is an integer and }a>0$

I understand that you repeatedly integrate by parts, but how does everything reduce to $\displaystyle \frac{k!}{a^{k+1}}$?

Thanks

\displaystyle \displaystyle \begin{align*} \int_0^{\infty}{ t^k \, e^{ -a \, t } \, dt } &= \frac{1}{ a^{k+1} } \int_0^{\infty}{ a^{k+1} \, t^k \, e^{-a\,t} \, dt } \\ &= \frac{1}{a^{k+1}} \int_0^{\infty}{ a \left( a\,t \right) ^k\, e^{-a\,t}\,dt} \\ &= \frac{1}{a^{k+1}}\int_0^{\infty}{ u^k\, e^{-u}\,du } \textrm{ after making the substitution } u = a\,t \implies du = a\,dt \\ &= \frac{1}{a^{k+1}} \int_0^{\infty}{ u^{(k+1) - 1} \, e^{-u}\,du } \\ &= \frac{1}{a^{k+1}}\,\Gamma{ \left( k + 1 \right) } \\ &= \frac{k!}{a^{k+1}} \textrm{ since } \Gamma{ (n)} = (n - 1)! \textrm{ if } n \in \mathbf{Z} \end{align*}
• Jun 26th 2013, 01:59 AM
downthesun01
Re: Integration question
Here's what I have after integrating by parts 10 times:

$\displaystyle \frac{-t^{k}}{a}e^{-at}+\frac{kt^{k-1}}{a^2}e^{-at}-\frac{k(k-1)t^{k-2}}{a^3}e^{-at}-...+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)(k-7)(k-8)t^{k-9}}{a^{10}}e^{-at}$
• Jun 26th 2013, 02:05 AM
downthesun01
Re: Integration question
Thank you. Where does the gamma come from?

Nevermind, I looked up the gamma function.
• Jun 26th 2013, 02:10 AM
Prove It
Re: Integration question
It's the Gamma Function, which is the continuous function you get if you interpolate the Factorial function so it is defined over most numbers instead of just the positive integers.
• Jun 26th 2013, 03:45 AM
JJacquelin
Re: Integration question
Quote:

Originally Posted by downthesun01
Here's what I have after integrating by parts 10 times:

$\displaystyle \frac{-t^{k}}{a}e^{-at}+\frac{kt^{k-1}}{a^2}e^{-at}-\frac{k(k-1)t^{k-2}}{a^3}e^{-at}-...+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)(k-7)(k-8)t^{k-9}}{a^{10}}e^{-at}$

You are on the good track !
But in the result of each intagration by part , there is a remaining integral (to be used at the next step). Where is the last remaining integral ?
You did it 10 times. OK. But what happend before, if k=4 for example ?
And do not forget that the limits of the integral are t=0 and t=infinity. Where are they applied into your result ?