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OpenStudy (amorfide):
\[\int\limits_{0}^{\infty} (e^{-t}t^{n})dt\]
OpenStudy (amorfide):
Apparently the answer is n! if that helps, idk how to get to that
OpenStudy (eliesaab):
You have to get a recurrence formula, or do integration by part n times.
sam (.sam.):
What's this? The answer is a gamma function
OpenStudy (amorfide):
I have to do it via integration by parts, and I have no idea how to succeed
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OpenStudy (amorfide):
Would I let u=t^n, or u=e^-t
OpenStudy (eliesaab):
Yes
OpenStudy (amorfide):
wow, which one???
OpenStudy (eliesaab):
If you call
\[
I_n=\int_0^\infty t^n e^{-t}dt
\]
OpenStudy (eliesaab):
When you do the first integration by part, you get
\[
I_n=n I_{n-1}
\]
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OpenStudy (eliesaab):
You can see where the n! would come from
OpenStudy (amorfide):
oh!
OpenStudy (amorfide):
wait
OpenStudy (amorfide):
\[u=t^{n}\]
\[\frac{ du }{ dt } = nt^{n-1}\]
\[\frac{ dv }{ dt }= e^{-t}\]
\[v=-e^{-t}\]
\[[-e^{-t}t^{n}] + n \int\limits_{}^{} e^{-t}t^{n-1}\]
the left hand side will always become 0, meaning you will end up with n(n-1)(n-2)...etc
thanks!