gamma integral

**phycdude** · November 7th 2011, 12:29 AM

i am trying to express the following in form of a gamma integral but so far i have been unsuccessful:
$- \int_0^1 x^k ln(x)$
some of the substitutions i have tried are: $x = \frac{1}{t}$ , $=t$ , $e^{-t}$ , $= log(\frac{1}{t})$ , $= e^{-\frac{1}{t}}$ and even $x^k = t$ , any pointer would be appreciated.
i am tring to prove the following result:
$-\int_0^1 x^k ln(x) = \frac{1}{(k+1)^2}$

edit i omitted the variable of integration so:
$- \int_0^1 x^k ln(x) dx$
for the first, and
$-\int_0^1 x^k ln(x) dx = \frac{1}{(k+1)^2}$
for the last integrals respectively.

**FernandoRevilla** · November 7th 2011, 01:11 AM

Originally Posted by phycdude

i am tring to prove the following result:
$-\int_0^1 x^k ln(x) = \frac{1}{(k+1)^2}$

Consider $I(k)=\int_{0}^1x^{k}\;dx=\dfrac{1}{k+1}$ . Using parametric derivation, $I'(k)=\int_{0}^1x^{k}\ln x\;dx$ . Conclude.

**phycdude** · November 7th 2011, 01:59 AM

i'd love to say i understand you method, but i do not, that said, im working on a topic in mathematicsl physics (gamma fuctions, beta fucntions and such), hence why i'm trying to convert this problem into the general form of gamma a function.

Originally Posted by FernandoRevilla

Consider $I(k)=\int_{0}^1x^{k}\;dx=\dfrac{1}{k+1}$ . Using parametric derivation, $I'(k)=\int_{0}^1x^{k}\ln x\;dx$ . Conclude.

edit:
see the attached scan of arfken and weber i think, sixth ed.

**FernandoRevilla** · November 7th 2011, 02:55 AM

In such a case, using the substitution $u=-(k+1)\log x$ you'll obtain:

$-\int_0^1x^k\ln x\;dx=\ldots=\frac{1}{(k+1)^2}\int_0^{+\infty}e^{-u}u\;du=\frac{\Gamma (2)}{(k+1)^2}=\frac{1}{(k+1)^2}$

**phycdude** · November 7th 2011, 03:19 AM

Thats brilliant, thanks

Originally Posted by FernandoRevilla

In such a case, using the substitution $u=-(k+1)\log x$ you'll obtain:

$-\int_0^1x^k\ln x\;dx=\ldots=\frac{1}{(k+1)^2}\int_0^{+\infty}e^{-u}u\;du=\frac{\Gamma (2)}{(k+1)^2}=\frac{1}{(k+1)^2}$

**chisigma** · November 10th 2011, 11:04 AM

Originally Posted by phycdude

i am trying to express the following in form of a gamma integral but so far i have been unsuccessful:
$- \int_0^1 x^k ln(x)$
some of the substitutions i have tried are: $x = \frac{1}{t}$ , $=t$ , $e^{-t}$ , $= log(\frac{1}{t})$ , $= e^{-\frac{1}{t}}$ and even $x^k = t$ , any pointer would be appreciated.
i am tring to prove the following result:
$-\int_0^1 x^k ln(x) = \frac{1}{(k+1)^2}$

Integration by parts permits an elementary solution...

$\int_{0}^{1}x^{k}\ \ln x\ dx = |\frac{x^{k+1}}{k+1}\ \ln x|_{0}^{1} - \frac{1}{k+1}\ \int_{0}^{1} x^{k}\ dx = - \frac{1}{(k+1)^{2}}$ (1)

Kind regards

$\chi$ $\sigma$

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