problem of the week: int [0, 1] [ln(x) / (x +1)] dx

[galactus]

I will post a problem of the week. How about a tricky integral for those of you who enjoy integration for kicks . I do .

Not that it's practical. Just fun puzzles to solve.

\(\displaystyle \int_{0}^{1}\frac{ln(x)}{x+1}dx\)

If no one bites in a few days I will assume lack of interest and delete it. OK?.

 

[Deleted member 4993]

Re: problem of the week?.

I will post a problem of the week. How about a tricky integral for those of you who enjoy integration for kicks . I do .

Not that it's practical. Just fun puzzles to solve.

\(\displaystyle \int_{0}^{1}\frac{ln(x)}{x+1}dx\)

If no one bites in a few days I will assume lack of interest and delete it. OK?.

ln(x) is not defined at x = 0.

 

[galactus]

Re: problem of the week?.

True, but this has a definite solution.

 

[chivox]

Re: problem of the week?.

I will post a problem of the week. How about a tricky integral for those of you who enjoy integration for kicks . I do .

Not that it's practical. Just fun puzzles to solve.

\(\displaystyle \int_{0}^{1}\frac{ln(x)}{x+1}dx\)

If no one bites in a few days I will assume lack of interest and delete it. OK?.

ln(x) is not defined at x = 0.

I'm also having a problem with the interval, but I enjoy this kind of thing... a place for the adults to talk. Don't take it down.

... Question: Is the integral defined at x = 0? I only ask because I think the integral is

\(\displaystyle ln(x) \cdot ln(x+1) + Li_2(-x)\)

Ahh, the polylogarithm, Jonquière's function, but ln(x) remains. This is too experimental for me, and I don't know how to go any further, so I leave that to the real mathematicians. Eddie? Where are you when we need you? Definition:

\(\displaystyle Li_n(z) = \sum_{k=1}^\infty \frac{z^k}{k^n}\)

 

[galactus]

Thanks for the input, chivox, but no need for that. One can do this by using series and a little parts.

Since it is a problem of the WEEK, I will leave it up one week to see what interest it gathers. I like to see different methods. I have mine(A portion of which appears below), but others are nice to see.

Here is a hint:

\(\displaystyle \int_{0}^{1}\frac{ln(x)}{x+1}dx=\int_{0}^{1}ln(x)\sum_{k=0}^{\infty}(-1)^{k}x^{k}dx=\sum_{k=0}^{\infty}(-1)^{k}\cdot \int_{0}^{1}x^{k}ln(x)dx\)

Well, since no one is going to bite, I may as well show my method.

Starting where I left off:

Using parts:

\(\displaystyle =\sum_{k=0}^{\infty}(-1)^{k}\left[\frac{xln(x)}{k+1}-\frac{x}{(k+1)^{2}}x^{k}\right]\)

Now, using the limits of integration 0 to 1, we get it whittled down to:

\(\displaystyle \sum_{k=0}^{\infty}(-1)^{k}\frac{-1}{(k+1)^{2}}\)

\(\displaystyle =\sum_{k=1}^{\infty}\frac{-(-1)^{k-1}}{k^{2}}\)

Now, we can use an identity related to the zeta function: \(\displaystyle (1-2^{1-x})\zeta(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^{x}}\)

\(\displaystyle -(1-2^{-1})\zeta(2)=-(1-\frac{1}{2})\frac{{\pi}^{2}}{6}=\frac{-{\pi}^{2}}{12}\)

There we have it:

\(\displaystyle \boxed{\int_{0}^{1}\frac{ln(x)}{x+1}dx=\frac{-{\pi}^{2}}{12}}\)

Knowledge of some series identities can be helpful when doing these obscure integrals. Zeta, beta, gamma often come into play.