SOLVED Riemann Integrable?

Jun 2010
205
16
I am asked to show that \(\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \frac{1}{(n + p)^{2}}\) for p > 0. I have a lot of it so far, and am basically on the cusp.

\(\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx\). Now the obvious calculus thing to do is integrate by parts and I'm done. But to do that, assuming I'm treating this like a Riemann integral, I have to evaluate \(\displaystyle log \Bigl( \frac{1}{x} \Bigr) \frac{x^{n + p}}{n + p}\) from 0 to 1. Well okay, 1 is easy. Log of 0 isn't defined, though, and I'm not sure how to do a proofy demonstration that this is whole thing is zero. So that leads me to think that maybe I shouldn't be treating this like a Riemann integral and instead should find some sequence of step functions which approaches what I'm trying to integrate, i.e. to argue it as a Lebesgue integral. But finding a sequence of step functions which approaches \(\displaystyle x^{n + p - 1} log(1/x)\) seems daunting. Ideas?
 

TheEmptySet

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Feb 2008
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I am asked to show that \(\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \frac{1}{(n + p)^{2}}\) for p > 0. I have a lot of it so far, and am basically on the cusp.

\(\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx\). Now the obvious calculus thing to do is integrate by parts and I'm done. But to do that, assuming I'm treating this like a Riemann integral, I have to evaluate \(\displaystyle log \Bigl( \frac{1}{x} \Bigr) \frac{x^{n + p}}{n + p}\) from 0 to 1. Well okay, 1 is easy. Log of 0 isn't defined, though, and I'm not sure how to do a proofy demonstration that this is whole thing is zero. So that leads me to think that maybe I shouldn't be treating this like a Riemann integral and instead should find some sequence of step functions which approaches what I'm trying to integrate, i.e. to argue it as a Lebesgue integral. But finding a sequence of step functions which approaches \(\displaystyle x^{n + p - 1} log(1/x)\) seems daunting. Ideas?
This should get you started

\(\displaystyle \int_{0}^{1}\frac{x^{p-1}}{1-x}\log\left( \frac{1}{x}\right)dx=\int_{0}^{1}\int_{x}^{1}\frac{1}{s}\cdot \frac{x^{p-1}}{1-x}dsdx\)

Now use Fubini's theorem to switch the order of integration to get

\(\displaystyle \int_{0}^{1}\int_{0}^{s}\frac{1}{s}\cdot \frac{x^{p-1}}{1-x}dxds\)

Now just use your trick and express \(\displaystyle \frac{1}{1-x}\) as geometric series and integrate.
 
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Jun 2010
205
16
I'm afraid I can't go that route since I don't have Fubini's Theorem or anything about double-integrals.
 
Jun 2010
205
16
... However, a similar idea might be to build the step-functions which approximate this integral on intervals [s, 1] where s is ever closer to 0 and the step functions are constant on smaller partitions as the sequence approaches the function. I'm toying with this idea right now.

Update:

How does this sound? I create the sequence of functions defined by \(\displaystyle f_{i} = \begin{cases}log \frac{1}{x} \text{ if } \frac{1}{i + 1} \leq x \leq 1 \\ 0 \text{ if } x > \frac{1}{i + 1} \end{cases}\). Obviously this sequence is increasing and approaching \(\displaystyle log \frac{1}{x}\), and each individual function is Riemann integrable and equal to... Hm. I guess I'm stuck here, I don't know how to integrate log yet. So this looks like the wrong way to go.
 
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Bruno J.

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Jun 2009
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Here's another idea not using Fubini's theorem. In the integral \(\displaystyle \int_0^1 x^{n+p-1}\log(x^{-1})dx\), change the variable to \(\displaystyle u=x^{-1}\). You get, assuming \(\displaystyle p>0\), \(\displaystyle \int_1^\infty u^{-n-p-1} \log u du = \int_1^\infty \log u\: d\left(\frac{u^{-n-p}}{-n-p}\right) = \log u\frac{u^{-n-p}}{-n-p} \Big |^{\infty}_1 - \int_1^\infty \frac{u^{-n-p-1}}{-n-p}du = \frac{1}{(-n-p)^2}\).
 
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Jose27

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Apr 2009
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I am asked to show that \(\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \frac{1}{(n + p)^{2}}\) for p > 0. I have a lot of it so far, and am basically on the cusp.

\(\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx\). Now the obvious calculus thing to do is integrate by parts and I'm done. But to do that, assuming I'm treating this like a Riemann integral, I have to evaluate \(\displaystyle log \Bigl( \frac{1}{x} \Bigr) \frac{x^{n + p}}{n + p}\) from 0 to 1. Well okay, 1 is easy. Log of 0 isn't defined, though, and I'm not sure how to do a proofy demonstration that this is whole thing is zero. So that leads me to think that maybe I shouldn't be treating this like a Riemann integral and instead should find some sequence of step functions which approaches what I'm trying to integrate, i.e. to argue it as a Lebesgue integral. But finding a sequence of step functions which approaches \(\displaystyle x^{n + p - 1} log(1/x)\) seems daunting. Ideas?
I think if p>1 your function has a continous extension to [0,1] so integration by parts would be justified. For the other cases (where the function blows up near 0) I think the question would be if \(\displaystyle \ln \left( \frac{1}{x} \right)\) is integrable.
 
Jun 2010
205
16
Here's another idea not using Fubini's theorem. In the integral \(\displaystyle \int_0^1 x^{n+p-1}\log(x^{-1})dx\), change the variable to \(\displaystyle u=x^{-1}\). You get, assuming \(\displaystyle p>0\), \(\displaystyle \int_1^\infty u^{-n-p-1} \log u du = \int_1^\infty \log u\: d\left(\frac{u^{-n-p}}{-n-p}\right) = \log u\frac{u^{-n-p}}{-n-p} \Big |^{\infty}_1 - \int_1^\infty \frac{u^{-n-p-1}}{-n-p}du = \frac{1}{(-n-p)^2}\).
This is, I think, a more elegant way of doing essentially what I just figured out how to do. Thank you all!