Hint at solving an integral, please?

**Pranas** · April 6th 2011, 12:18 PM

Hello.

$\displaystyle \[I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}} \cdot \ln (x)}}{{x + 1}}dx} \]$

and

$\displaystyle \[\frac{d}{{dp}}I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}} \cdot {{(\ln (x))}^2}}}{{x + 1}}dx} \]$

seems useless to me.
So I tried substituting

$\displaystyle \[\ln (x + 1) = u\]$

which apparently gave me

$\displaystyle \[I(p) = \int\limits_0^{ + \infty } {{e^{(u - 1)(p - 1)}}(u - 1)du} \]$

I am not sure if this is really true. Moreover, I don't seem to be able to solve them (I actually expect to get an Euler's gamma function somehow).

**derdack** · April 6th 2011, 12:57 PM

Try this way:

u=(x^(p-1))/(x+1) and dv=ln(x)dx. Find du and v, then you will get I(p)=u*v-Integrate[(v*du)]

**Pranas** · April 7th 2011, 07:49 AM

Originally Posted by derdack

Try this way:

u=(x^(p-1))/(x+1) and dv=ln(x)dx. Find du and v, then you will get I(p)=u*v-Integrate[(v*du)]

I get nothing elegant coming out of it, just more complicated problems.

Answer is supposed to be $\displaystyle \[\frac{{ - {\pi ^2} \cdot \cos (\pi p)}}{{{{(\sin (\pi p))}^2}}}\]$

which leaves me almost certain this can be achieved by making Euler's beta function

Edit: It took some hours of pure thinking, but satisfaction is well worth it.

$\displaystyle \[I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}}\cdot\ln (x)}}{{x + 1}}dx} = \frac{d}{{dp}}\int\limits_0^{ + \infty } {\left( {\int {\frac{{{x^{p - 1}}\cdot\ln (x)}}{{x + 1}}dp} } \right)dx} = \frac{d}{{dp}}\int\limits_0^{ + \infty } {\frac{{{x^{p - 1}}}}{{x + 1}}dx} \]$

Integral is already a form of Euler's beta function and known manipulations can be done:

$\displaystyle \[\int\limits_0^{ + \infty } {\frac{{{x^{p - 1}}}}{{x + 1}}dx} = {\rm B}\left( {p;1 - p} \right) = \frac{\pi }{{\sin (\pi p)}}\]$

Hence

$\displaystyle \[I(p) = \frac{d}{{dp}}\frac{\pi }{{\sin (\pi p)}} = \frac{{ - {\pi ^2} \cdot \cos (\pi p)}}{{{{(\sin (\pi p))}^2}}}\]$

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