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I have a problem with solving this integration problem:

$\int\limits_0^b x^{n-1}e'(t(x))dx$

I tried to solve it through partial integration by setting:

$f'(x) = e'(t(x)), f(x) = e(t(x)), g(x) = x^{n-1}, g'(x) = (n-1)x^{n-2}$

then I got this:

$\left[e(t(x))x^{n-1}\right]_0^b - \int\limits_0^b e(t(x))(n-1)x^{n-2}dx$

Now my idea was to do partial integration on the last expression again but got something really strange. Anyone any idea?


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If $e(t(x))=\mathrm{e}^{t(x)}=\exp(t(x))$ then your integral may not be expressible in terms of elementary functions (e.g. take $t(x)=x^{2}$). –  Shaktal Jun 17 '13 at 19:05
What is known about $t(x)$ - could you narrow it down perhaps? –  gt6989b Jun 17 '13 at 19:10
In a very simple case, this looks like a twist on the Gamma function (link here: en.wikipedia.org/wiki/Gamma_function): $$ \Gamma(n) = \int_0^\infty x^{n-1} e^{-x} dx $$ –  gt6989b Jun 17 '13 at 19:13
No e is unfortunately not an exponential function. I am currently doing the derivation for this paper: personal.psu.edu/amk17/TimeIsMoney_onlineversion.pdf. e is an enjoyment term and it is assumed that $e(t) = et$. The solution of this problem should somehow be: $e \frac{b^n}{n}$ –  Dat Tran Jun 17 '13 at 19:19

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