Logarithm integral

**Jem** · January 1st 2008, 07:48 AM

Evaluate $\int_0^1\frac{\ln(x+1)}x\,dx$ .

How many methods you know to solve it?

**ThePerfectHacker** · January 1st 2008, 08:19 AM

Originally Posted by Jem

Evaluate $\int_0^1\frac{\ln(x+1)}x\,dx$ .

How many methods you know to solve it?

$\ln (x+1) = \sum_{n=1}^{\infty} \frac{(-1)^nx^n}{n}$ .

Thus, $\frac{\ln (x+1)}{x} = \sum_{n=1}^{\infty} \frac{(-1)^n x^{n-1}}{n}$ .

This means, $\int_0^1 \left( \sum_{n=1}^{\infty} \frac{(-1)^n x^{n-1}}{n}\right) dx = \sum_{n=1}^{\infty} \left( \int_0^1 \frac{(-1)^{n+1} x^{n-1}}{n} dx \right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}= \frac{\pi^2}{12}$

**galactus** · January 1st 2008, 10:12 AM

This particular integral is related to what is known as the 'dilogarithm'.

$dilog(x)=\int_{1}^{x}\frac{ln(t)}{t-1}dt$

In our case, $\int_{0}^{1}\frac{ln(x+1)}{x}dx=\int_{1}^{2}\frac{ ln(x)}{x-1}dx = -dilog(x+1) = \frac{{\pi}^{2}}{12}$

It seems to me there is also a unit disk reference with the dilog. It says for n=2 and z in the unit disk, then

$Li_{2}(z)=\sum_{k=1}^{\infty}\frac{z^{k}}{k^{2}}$

i.e, if z=1, then we have $\frac{{\pi}^{2}}{6}$

There is also a trilogarithm.

**bobak** · January 1st 2008, 10:33 AM

what is dilog galactus?

**galactus** · January 1st 2008, 11:04 AM

Do a google to find out more. It is a cool function with lotsa implications.

For instance, it has the series: $-x-\frac{x^{2}}{4}-\frac{x^{3}}{9}-.....-\frac{x^{n}}{n^{2}}$

Also, $Li_{2}(z)=\int_{z}^{0}\frac{ln(1-t)}{t}dt$

In the event z = -1, then we have what PH posted.

Notice how the Basel sum is involved here. $\sum_{k=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2} }{6}$

I noticed the similarity between this and the posters integral and thought there was a connection. I am sure Wiki or Mathworld has something about it.

The dilog is the polylog with n=2. There are others. As I mentioned, a trilog.

It has implications to gamma and other things. Research it. You'll find it cool.

**Krizalid** · January 18th 2008, 08:31 AM

Originally Posted by ThePerfectHacker

$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}= \frac{\pi^2}{12}$

$\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }} {{n^2 }}} = 1 - \frac{1} {{2^2 }} + \frac{1} {{3^2 }} - \frac{1} {{4^2 }} + \cdots .$

This last is equal to

$\left( {1 + \frac{1} {{2^2 }} + \frac{1} {{3^2 }} + \frac{1} {{4^2 }} + \cdots } \right) - 2\left( {\frac{1} {{2^2 }} + \frac{1} {{4^2 }} + \cdots } \right).$

Since $2\left( {\frac{1} {{2^2 }} + \frac{1} {{4^2 }} + \cdots } \right) = \frac{2} {{2^2 }}\left( {1 + \frac{1} {{2^2 }} + \cdots } \right),$

we have $\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }} {{n^2 }}} = \sum\limits_{n = 1}^\infty {\frac{1} {{n^2 }}} - \frac{1} {2}\sum\limits_{n = 1}^\infty {\frac{1} {{n^2 }}} = \frac{{\pi ^2 }} {6} - \frac{{\pi ^2 }} {{12}} = \frac{{\pi ^2 }} {{12}}.$

**ThePerfectHacker** · January 18th 2008, 08:35 AM

Originally Posted by Krizalid

$\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }} {{n^2 }}} = 1 - \frac{1} {{2^2 }} + \frac{1} {{3^2 }} - \frac{1} {{4^2 }} + \cdots .$

This last is equal to

$\left( {1 + \frac{1} {{2^2 }} + \frac{1} {{3^2 }} + \frac{1} {{4^2 }} + \cdots } \right) - 2\left( {\frac{1} {{2^2 }} + \frac{1} {{4^2 }} + \cdots } \right).$

Since $2\left( {\frac{1} {{2^2 }} + \frac{1} {{4^2 }} + \cdots } \right) = \frac{2} {{2^2 }}\left( {1 + \frac{1} {{2^2 }} + \cdots } \right),$

we have $\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }} {{n^2 }}} = \sum\limits_{n = 1}^\infty {\frac{1} {{n^2 }}} - \frac{1} {2}\sum\limits_{n = 1}^\infty {\frac{1} {{n^2 }}} = \frac{{\pi ^2 }} {6} - \frac{{\pi ^2 }} {{12}} = \frac{{\pi ^2 }} {{12}}.$

That is the standard way this identity is derived from $\zeta (2)$ .

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