# Thread: Riemann Integrable?

1. ## Riemann Integrable?

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

$\displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx$. 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 $log \Bigl( \frac{1}{x} \Bigr) \frac{x^{n + p}}{n + p}$ 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 $x^{n + p - 1} log(1/x)$ seems daunting. Ideas?

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

$\displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx$. 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 $log \Bigl( \frac{1}{x} \Bigr) \frac{x^{n + p}}{n + p}$ 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 $x^{n + p - 1} log(1/x)$ seems daunting. Ideas?
This should get you started

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

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

$\int_{0}^{1}\int_{0}^{s}\frac{1}{s}\cdot \frac{x^{p-1}}{1-x}dxds$

Now just use your trick and express $\frac{1}{1-x}$ as geometric series and integrate.

3. I'm afraid I can't go that route since I don't have Fubini's Theorem or anything about double-integrals.

4. ... 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 $f_{i} = \begin{cases}log \frac{1}{x} \text{ if } \frac{1}{i + 1} \leq x \leq 1 \\ 0 \text{ if } x > \frac{1}{i + 1} \end{cases}$. Obviously this sequence is increasing and approaching $log \frac{1}{x}$, 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.

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

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

$\displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx$. 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 $log \Bigl( \frac{1}{x} \Bigr) \frac{x^{n + p}}{n + p}$ 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 $x^{n + p - 1} log(1/x)$ 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 $\ln \left( \frac{1}{x} \right)$ is integrable.

7. Originally Posted by Bruno J.
Here's another idea not using Fubini's theorem. In the integral $\int_0^1 x^{n+p-1}\log(x^{-1})dx$, change the variable to $u=x^{-1}$. You get, assuming $p>0$, $\int_1^\infty u^{-n-p-1} \log u du = \int_1^\infty \log u\: d\left(\frac{u^{-n-p}}{-n-p}\right) = \log u\frac{u^{-n-p}}{-n-p} \Big |^{\infty}_1 - \int_1^\infty \frac{u^{-n-p-1}}{-n-p}du = \frac{1}{(-n-p)^2}$.
This is, I think, a more elegant way of doing essentially what I just figured out how to do. Thank you all!