1. ## Integral-ln

Calculate :

2. Originally Posted by dhiab
Calculate :

Is it a definite integral actually ?

If we replace x by $\displaystyle e^t$
then $\displaystyle dx = e^t dt$

the integral $\displaystyle = \int e^t \sqrt{t} ~dt$

I am afraid that i dont have confidence to solve it !

If the u. limit is 1 and l. limit is 0 , then this integral

$\displaystyle \int_0^1 \sqrt{-\ln{x}}~dx ~~$ can be written as

$\displaystyle \int_0^{\infty} e^{-t} \sqrt{t} ~dt = \Gamma(3/2) = \frac{\sqrt{\pi}}{2}$

3. Originally Posted by dhiab
Calculate :
According to Mathematica, the answer is

$\displaystyle x\sqrt{\ln{x}} - \frac{1}{2}\sqrt{\pi}\,\textrm{erfi}(\sqrt{\ln{x}} )$

Erfi -- from Wolfram MathWorld

4. There is an identity which states:

$\displaystyle \int ln^{n}(x)dx=(-1)^{n}\cdot x\sum_{k=0}^{n}\frac{n!}{k!}(-ln(x))^{k}$

This would be interesting to prove.

Also, since $\displaystyle Ex(x,n)=\sum_{k=0}^{n}\frac{x^{k}}{k!}$,

the above can be written as $\displaystyle (-1)^{n}n!xEx(-ln(x),n)$

and $\displaystyle Ex(x,n)=\frac{e^{x}\Gamma(n+1,x)}{\Gamma(n+1)}$

where $\displaystyle {\Gamma}(x,n)$ is the incomplete gamma function.

Just some food for thought.

5. Originally Posted by galactus
There is an identity which states:

$\displaystyle \int ln^{n}(x)dx=(-1)^{n}\cdot x\sum_{k=0}^{n}\frac{n!}{k!}(-ln(x))^{k}$

This would be interesting to prove.
I might try letting $\displaystyle x = e^t$ (as mention earlier by SP) giving

$\displaystyle \int t^n e^t\,dt$
then try repeated integration by parts.