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- Thread starter bincybn
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- Feb 5, 2012

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Dear All,

\(\displaystyle \int_{0}^{1}x*\left(\left(-ln(x)\right)^{k}\right)*\left(1-x\right)^{(N-1)}dx \), where k is an odd no. N >=2.

regards,

Bincy

Hi bincybn,

Your integral cannot be expressed in terms of elementary functions. However using Maxima I found that the answer is,

\[\displaystyle \int_{0}^{1}x((-ln(x))^{k}\left(1-x\right)^{(N-1)}dx=(-1)^k\left( \left. \frac{{d}^{k}}{d\,{x}^{k}}\,\beta\left(N,x\right) \right|_{x=2}\right)~~~~\mbox{for }k\in\mathbb{Z^+}\]

\(\beta\) is the Beta function.

Since the integral is set up like the beta function, maybe he was supposed to obtain the solution by induction.Hi bincybn,

Your integral cannot be expressed in terms of elementary functions. However using Maxima I found that the answer is,

\[\displaystyle \int_{0}^{1}x((-ln(x))^{k}\left(1-x\right)^{(N-1)}dx=(-1)^k\left( \left. \frac{{d}^{k}}{d\,{x}^{k}}\,\beta\left(N,x\right) \right|_{x=2}\right)~~~~\mbox{for }k\in\mathbb{Z^+}\]

\(\beta\) is the Beta function.

- Feb 5, 2012

- 1,621

Hi dwsmith,Since the integral is set up like the beta function, maybe he was supposed to obtain the solution by induction.

Yes, maybe. But without knowing the answer it is impossible to use the induction method. Isn't?

Gamma function [tex]\Gamma (z) = \int_{0}^{\infty} e^{-t} t^{z-1} dt [/tex]

[tex]\int_{0}^{1} x((- \ln x )^k) (1-x)^{n-1} dx [/tex]

Let [tex] u = - \ln x \Rightarrow x = e^{-u} \Rightarrow dx = - e^{-u} du [/tex]

[tex]x = 0 \rightarrow u = \infty , x = 1 \rightarrow u = 0 [/tex]

[tex]\int_{\infty}^{0} e^{-u} (u^k )(1-e^{-u})^n(-e^{-u}) du [/tex]

[tex]\int_{0}^{\infty} e^{-2u} (u^k )(1-e^{-u})^n du [/tex]

Note that

[tex](1-e^{-u})^n = \sum_{i=0}^{n} \dbinom{n}{i} (-1)^i (e^{-u})^i [/tex]

and [tex]\int_{0}^{\infty} e^{-at} t^{z-1} dt = \frac{\Gamma (z)}{a} [/tex]

[tex]\int_{0}^{\infty} e^{-2u} (u^k )(1-e^{-u})^n du = \sum_{i=0}^{n} (-1)^i \int_{0}^{\infty} (u^k)(e^{-u(2+i)}) du [/tex]

[tex]\sum_{i=0}^{n} (-1)^i \int_{0}^{\infty} (u^k)(e^{-u(2+i)}) du =

\sum_{i=0}^{\infty}\frac{(-1)^i\Gamma (k+1)}{2+i}[/tex]

I solved it for n instead of n-1

[tex]\int_{0}^{1} x((- \ln x )^k) (1-x)^{n-1} dx [/tex]

Let [tex] u = - \ln x \Rightarrow x = e^{-u} \Rightarrow dx = - e^{-u} du [/tex]

[tex]x = 0 \rightarrow u = \infty , x = 1 \rightarrow u = 0 [/tex]

[tex]\int_{\infty}^{0} e^{-u} (u^k )(1-e^{-u})^n(-e^{-u}) du [/tex]

[tex]\int_{0}^{\infty} e^{-2u} (u^k )(1-e^{-u})^n du [/tex]

Note that

[tex](1-e^{-u})^n = \sum_{i=0}^{n} \dbinom{n}{i} (-1)^i (e^{-u})^i [/tex]

and [tex]\int_{0}^{\infty} e^{-at} t^{z-1} dt = \frac{\Gamma (z)}{a} [/tex]

[tex]\int_{0}^{\infty} e^{-2u} (u^k )(1-e^{-u})^n du = \sum_{i=0}^{n} (-1)^i \int_{0}^{\infty} (u^k)(e^{-u(2+i)}) du [/tex]

[tex]\sum_{i=0}^{n} (-1)^i \int_{0}^{\infty} (u^k)(e^{-u(2+i)}) du =

\sum_{i=0}^{\infty}\frac{(-1)^i\Gamma (k+1)}{2+i}[/tex]

I solved it for n instead of n-1

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To Amer: Thanks for ur method. But some mistakes are here and there like.

which is in fact [tex]\int_{0}^{\infty} e^{-at} t^{z-1} dt = \frac{\Gamma (z)}{a^z} [/tex][tex]\int_{0}^{\infty} e^{-at} t^{z-1} dt = \frac{\Gamma (z)}{a} [/tex]

regards,

Bincy