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