## UnkleRhaukus 2 years ago $\int\limits_0^1u^x(\ln u)^n\,\mathrm du\qquad n\in\mathbb Z>0$

1. UnkleRhaukus

\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*}

2. slaaibak

woah.

3. UnkleRhaukus

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

4. UnkleRhaukus

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

5. sirm3d

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

6. UnkleRhaukus

im not sure what i would do with such a result

7. sirm3d

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

8. sirm3d

$\Large \lim_{u\rightarrow 0+}u \ln u = 0$

9. sirm3d

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

10. UnkleRhaukus

0^0=0?

11. slaaibak

it's u^x, not u^n

12. sirm3d

$$n$$ n is a positive integer so $$0^n=0$$

13. sirm3d

oh, its $$x$$, not $$n$$.

14. sirm3d

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

15. sirm3d

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

16. sirm3d

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

17. sirm3d

hmm. my solution differs by a negative sign.

18. UnkleRhaukus

hmm

19. sirm3d

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.

20. experimentX

|dw:1357579412038:dw| looks like this could have been shorter.

21. UnkleRhaukus

Yeah thats neat too