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\[\int\limits_0^1u^x(\ln u)^n\,\mathrm du\qquad n\in\mathbb Z>0\]

Mathematics
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\[\begin{align*} &\int\limits_0^1u^x(\ln u)^n\,\mathrm du\qquad&n\in\mathbb Z>0\\ &\text{let } u =e^{-w}\\ &\mathrm du=-e^{-w}\mathrm dw\\ &u=0\rightarrow w=\infty\\ &u=1\rightarrow w=0\\ &=\int\limits_\infty^0e^{-xw}(-w)^n(-e^{-w})\mathrm dw\\ &=(-1)^{n+1}\int\limits_0^\infty w^ne^{-(x+1)w}\mathrm dw\\ &=(-1)^{n+1}\left[\left.\frac{w^ne^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty \frac{nw^{n-1}e^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\ &=(-1)^{n+1}\left[0+\frac{n}{x+1}\int\limits_0^\infty w^{n-1}e^{-(x+1)w}\mathrm dw\right]\\ &=\frac{n(-1)^{n+1}}{x+1}\int\limits_0^\infty w^{n-1}e^{-(x+1)w}\mathrm dw\\ &=\frac{n(-1)^{n+1}}{x+1}\left[\left.\frac{w^{n-1}e^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty (n-1)w^{n-2}\frac{e^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\ &=\frac{n(-1)^{n+1}}{x+1}\left[0+\frac{n-1}{x+1}\int\limits_0^\infty w^{n-2}e^{-(x+1)w}\mathrm dw\right]\\ &=\frac{n(n-1)(-1)^{n+1}}{(x+1)^2}\int\limits_0^\infty w^{n-2}e^{-(x+1)w}\mathrm dw\\ &=\frac{n(n-1)(-1)^{n+1}}{(x+1)^2}\left[\left.\frac{w^{n-2}w^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty\frac{(n-2)w^{n-3}w^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\ &=\frac{n(n-1)(-1)^{n+1}}{(x+1)^2}\left[0+\frac{n-2}{x+1}\int\limits_0^\infty w^{n-3}w^{-(x+1)w}\mathrm dw\right]\\ &=\frac{n(n-1)(n-2)(-1)^{n+1}}{(x+1)^3}\int\limits_0^\infty w^{n-3}w^{-(x+1)w}\mathrm dw\\ &=\qquad\vdots\\ \\ &=\frac{n!(-1)^{n+1}}{(x+1)^n} \end{align*}\]
woah.
i think i made a mistake some where, because the answer should be \[\boxed{=\dfrac{n!(-1)^{n}}{(x+1)^n}}\]

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Other answers:

ah yes there are three negatives on that line , \[=\int\limits_\infty^0e^{-xw}(-w)^n(-e^{-w})\mathrm dw\\ =(-1)^n\int\limits_\infty^0e^{-xw}w^n(-e^{-w})\mathrm dw\\ =-(-1)^n\int\limits_\infty^0e^{-xw}w^ne^{-w}\mathrm dw\\ =(-1)^n\int\limits^\infty_0e^{-xw}w^ne^{-w}\mathrm dw\\ =(-1)^{n}\int\limits_0^\infty w^ne^{-(x+1)w}\mathrm dw\\ =\quad\vdots\]
you can also use integration by parts, arriving at a reduction formula \[\large\int u^n(\ln u)^n \mathrm du=\frac{1}{n+1}u^{n+1}(\ln u)^n - \frac{n}{n+1}\int u^n(\ln u)^{n-1} \mathrm du,n\in \mathbb Z^{+}\]
im not sure what i would do with such a result
\[\left. u^{n+1}{(\ln u)^n} \right|\;_0^1=?\]
\[\Large \lim_{u\rightarrow 0+}u \ln u = 0\]
\[\Large u^{n+1}(\ln u)^n=u(u\ln u)^n=0(0)^n=0\]
0^0=0?
it's u^x, not u^n
\(n\) n is a positive integer so \(0^n=0\)
oh, its \(x\), not \(n\).
\[\int_0^1u^x(\ln u)^n\mathrm du=\frac{1}{x+1}u^{x+1-n}u^n(\ln u)^n-\frac{n}{x+1}\int_0^1u^x(\ln u)^{n-1}\mathrm du\] for as long as \(x+1-n>0\), the first term in the RHS is zero. THUS, \[\int_0^1u^x(\ln u)^n\mathrm du=-\frac{n}{x+1}\int_0^1u^x(\ln u)^{n-1}\mathrm du\] eventually, n-1=0 because of the reduction formula.
\[\int_0^1u^x(\ln u)^n \mathrm du=(-1)^{n-1}\frac{n(n-1)\cdots(2)}{\underbrace{(x+1)(x+1)\cdots(x+1)}_{n-1\text{ factors}}}\int_0^1 u^x(\ln u)^{1-1} \mathrm du\] now, the integral on the RHS is \[\frac{1}{x+1}\] and so the desired result \[\frac{n!}{(x+1)^n}\]
i mean \[(-1)^{n-1}\frac{n!}{(x+1)^n}\] as desired.
hmm. my solution differs by a negative sign.
hmm
if i had written \[\int u^x(\ln u)^n \mathrm du=(-1)^{1}\frac{n}{x+1}\int u^x(\ln u)^{n-1}\mathrm du\\=(-1)^2\frac{n}{x+1}\frac{n-1}{x+1}\int u^x (\ln u)^{n-2}\mathrm du\\\vdots\\=(-1)^n\frac{n(n-1)\cdots(2)}{(x+1)^{n-1}}\int_0^1 u^x (\ln u)^0 \mathrm du\] note that the exponent of (-1) and (ln u) add up to n, and the recursion is applied (n-1) times so the desired result can be concluded.
|dw:1357579412038:dw| looks like this could have been shorter.
Yeah thats neat too

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