# Thread: Proving the Integral Exists

1. ## Proving the Integral Exists

I am trying to prove that $\displaystyle \displaystyle \int^{1}_{0} \frac{x logx}{(1+x)^{2}}dx$ exists as a Lebesgue integral. I've tried evaluating the integral a couple ways, and if I kept pursuing these routes I might find an answer, but these methods are getting a little tortuous and I suspect this is not the best or intended answer. I have all of Levi's Theorems and the Lebesgue Dominated Convergence Theorem, and I sort of suspect that I'm supposed to derive the result from these or something like them, but I'm having a hard time seeing how that would go.

2. Well, one avenue you can take is that the Riemann integral exists. If a function is Riemann integrable, it is Lebesgue integrable.

3. I'm not sure how to show that this is Riemann integrable since, on the interval, log(x) isn't. But this suggestion did inspire another method of proof which I'm pretty sure works, so many thanks!

4. Mat be that the best way to demonstrate that a definite integral exists is its computation!... at this scope we start considerating that for $\displaystyle |x|<1$ is...

$\displaystyle \displaystyle \frac{1}{(1+x)^{2}} = - \frac{d}{dx}\ \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^{n} (n+1)\ x^{n}$ (1)

Now taking into account (1) and the general formula...

$\displaystyle \displaystyle \int_{0}^{1} x^{n}\ \ln x\ dx = - \frac{1}{(n+1)^{2}}$ (2)

... You obtain...

$\displaystyle \displaystyle \int_{0}^{1} \frac{x\ \ln x}{(1+x)^{2}}\ dx = \sum_{n=0}^{\infty} (-1)^{n}\ (n+1)\ \int_{0}^{1} x^{n+1}\ \ln x \ dx =$

$\displaystyle \displaystyle = - \sum_{n=0}^{\infty} (-1)^{n} \frac{n+1}{(n+2)^{2}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+2)^{2}} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+2} = \ln 2 - \frac{\pi^{2}}{12} = - .129319852864...$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Just to add, it's easy to see that the function has a continous extension to [0,1] so it's Riemann integrable there.

6. Is important to observe that the integral...

$\displaystyle \displaystyle \int_{0}^{1} x^{n}\ \ln x \ dx$ (1)

...for $\displaystyle n \ge 1$ exists as Riemann integral because the function is bounded in $\displaystyle [0,1]$. In particular for $\displaystyle n \ge 1$ is ...

$\displaystyle \displaystyle \lim_{x \rightarrow 0 +} x^{n}\ \ln x = 0$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$