Question

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

Answer 1

\[\begin{align*}
&\int\limits_0^1u^x(\ln u)^n\,\mathrm du\qquad&n\in\mathbb Z>0\\
&\text{let } u =e^{-w}\\
&\mathrm du=-e^{-w}\mathrm dw\\
&u=0\rightarrow w=\infty\\
&u=1\rightarrow w=0\\
&=\int\limits_\infty^0e^{-xw}(-w)^n(-e^{-w})\mathrm dw\\
&=(-1)^{n+1}\int\limits_0^\infty w^ne^{-(x+1)w}\mathrm dw\\
&=(-1)^{n+1}\left[\left.\frac{w^ne^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty \frac{nw^{n-1}e^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\
&=(-1)^{n+1}\left[0+\frac{n}{x+1}\int\limits_0^\infty w^{n-1}e^{-(x+1)w}\mathrm dw\right]\\
&=\frac{n(-1)^{n+1}}{x+1}\int\limits_0^\infty w^{n-1}e^{-(x+1)w}\mathrm dw\\
&=\frac{n(-1)^{n+1}}{x+1}\left[\left.\frac{w^{n-1}e^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty (n-1)w^{n-2}\frac{e^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\
&=\frac{n(-1)^{n+1}}{x+1}\left[0+\frac{n-1}{x+1}\int\limits_0^\infty w^{n-2}e^{-(x+1)w}\mathrm dw\right]\\
&=\frac{n(n-1)(-1)^{n+1}}{(x+1)^2}\int\limits_0^\infty w^{n-2}e^{-(x+1)w}\mathrm dw\\
&=\frac{n(n-1)(-1)^{n+1}}{(x+1)^2}\left[\left.\frac{w^{n-2}w^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty\frac{(n-2)w^{n-3}w^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\
&=\frac{n(n-1)(-1)^{n+1}}{(x+1)^2}\left[0+\frac{n-2}{x+1}\int\limits_0^\infty w^{n-3}w^{-(x+1)w}\mathrm dw\right]\\
&=\frac{n(n-1)(n-2)(-1)^{n+1}}{(x+1)^3}\int\limits_0^\infty w^{n-3}w^{-(x+1)w}\mathrm dw\\
&=\qquad\vdots\\
\\
&=\frac{n!(-1)^{n+1}}{(x+1)^n}
\end{align*}\]

Answer 2

woah.

Answer 3

i think i made a mistake some where, because the answer should be
\[\boxed{=\dfrac{n!(-1)^{n}}{(x+1)^n}}\]

Answer 4

ah yes there are three negatives on that line , \[=\int\limits_\infty^0e^{-xw}(-w)^n(-e^{-w})\mathrm dw\\
=(-1)^n\int\limits_\infty^0e^{-xw}w^n(-e^{-w})\mathrm dw\\
=-(-1)^n\int\limits_\infty^0e^{-xw}w^ne^{-w}\mathrm dw\\
=(-1)^n\int\limits^\infty_0e^{-xw}w^ne^{-w}\mathrm dw\\
=(-1)^{n}\int\limits_0^\infty w^ne^{-(x+1)w}\mathrm dw\\
=\quad\vdots\]

Answer 5

you can also use integration by parts, arriving at a reduction formula
\[\large\int u^n(\ln u)^n \mathrm du=\frac{1}{n+1}u^{n+1}(\ln u)^n - \frac{n}{n+1}\int u^n(\ln u)^{n-1} \mathrm du,n\in \mathbb Z^{+}\]

Answer 6

im not sure what i would do with such a result

Answer 7

\[\left. u^{n+1}{(\ln u)^n} \right|\;_0^1=?\]

Answer 8

\[\Large \lim_{u\rightarrow 0+}u \ln u = 0\]

Answer 9

\[\Large u^{n+1}(\ln u)^n=u(u\ln u)^n=0(0)^n=0\]

Answer 10

0^0=0?

Answer 11

it's u^x, not u^n

Answer 12

\(n\) n is a positive integer so \(0^n=0\)

Answer 13

oh, its \(x\), not \(n\).

Answer 14

\[\int_0^1u^x(\ln u)^n\mathrm du=\frac{1}{x+1}u^{x+1-n}u^n(\ln u)^n-\frac{n}{x+1}\int_0^1u^x(\ln u)^{n-1}\mathrm du\]
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.

Answer 15

\[\int_0^1u^x(\ln u)^n \mathrm du=(-1)^{n-1}\frac{n(n-1)\cdots(2)}{\underbrace{(x+1)(x+1)\cdots(x+1)}_{n-1\text{ factors}}}\int_0^1 u^x(\ln u)^{1-1} \mathrm du\]
now, the integral on the RHS is \[\frac{1}{x+1}\] and so the desired result \[\frac{n!}{(x+1)^n}\]

Answer 16

i mean \[(-1)^{n-1}\frac{n!}{(x+1)^n}\] as desired.

Answer 17

hmm. my solution differs by a negative sign.

Answer 18

hmm

Answer 19

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.

Answer 20

looks like this could have been shorter.