An interesting logarithm integral

In summary, the conversation revolves around an integral that was seen in another forum and the value that was returned by W|A. The thread is dedicated to solving the integral and any ideas are welcomed. It is suggested that the integral might require some dilogarithm identities and the first step could be integration by parts. There is a discussion about whether the "log" operator is base 10 or e, and it is determined to be the natural logarithm. Various approaches are discussed, including using dilogarithm identities and integration by parts. A formula for integrals of the form f(x) = \sum_{k=0}^{\infty} a_{k}\ x^{k} is mentioned, and it is applied to
  • #1
alyafey22
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MHB
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Here is an integral that I seen in another forum

\(\displaystyle \int^1_0 \frac{\log(x)\log(x+1)}{1-x} \approx -0.5 \)

That was a value return by W|A . This thread will be dedicated to try to solve the integral . Any ideas will always be welcomed .

The integral might require some dilogarithm identities . The first step might be integration by parts .
 
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  • #2
ZaidAlyafey said:
Here is an integral that I seen in another forum

\(\displaystyle \int^1_0 \frac{\log(x)\log(x+1)}{1-x} \approx -0.5 \)

That was a value return by W|A . This thread will be dedicated to try to solve the integral . Any ideas will always be welcomed .

The integral might require some dilogarithm identities . The first step might be integration by parts .
Looking at the dilogarithms I am lead to ask if the "log" operator is base 10 or e?

-Dan
 
  • #3
It is a based e logarithm . In complex analysis we always assume \(\displaystyle \log\) to be the natural logarithm , it looks better than \(\displaystyle \ln \,\, \) (Cool).
 
  • #4
Ok , I tried integration by parts

\(\displaystyle \int^1_0 \frac{\log(x) \log(1-x)}{1+x}\, dx +\int^1_0 \frac{\log(1+x) \log(1-x)}{x}\, dx\)

The first integral

I might start by the following

\(\displaystyle \int^1_0 \frac{x^{a-1}(1-x)^{b-1}}{1+x} = \int^1_0 \left( \sum_{n \geq 0 }(-x)^n \right) x^{a-1}(1-x)^{b-1} \, dx \)\(\displaystyle \sum_{n \geq 0 }(-1)^n \left( \int^1_0 x^{n+a-1}(1-x)^{b-1} \, dx \right) \)\(\displaystyle \int^1_0 \frac{x^{a-1}(1-x)^{b-1}}{1+x} =\Gamma(b) \sum_{n \geq 0 }(-1)^n \frac{\Gamma(n+a)}{\Gamma(n+a+b)} = \phi (a,b)\)

\(\displaystyle \int^1_0 \frac{x^{a-1} (1-x)^{b-1} \log(x) \log(1-x) }{1+x} = \frac{\partial }{\partial a} \left( \frac{\partial}{\partial b} \phi (a,b) \right)\)

\(\displaystyle \int^1_0 \frac{\log(x) \log(1-x) }{1+x} = \frac{\partial \phi (1,1) }{\partial a \, \partial b} \)

Our best bet is to find the sum

\(\displaystyle \sum_{n \geq 0 }(-1)^n \frac{\Gamma(n+a)}{\Gamma(n+a+b)}\)
 
  • #5
ZaidAlyafey said:
\(\displaystyle I= \int^1_0 \frac{\log(x)\log(x+1)}{1-x}\, dx \)

I have got a possibly better idea

let \(\displaystyle x = 1-t \)

\(\displaystyle I=\int^1_0 \frac{\log(1-t)\log(2-t)}{t}\, dt = \int^1_0 \frac{\log(1-t)\left ( \log(1-\frac{t}{2}) + \log(2) \right)}{t}\, dt \)

\(\displaystyle I=\int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt +\log(2)\int^1_0 \frac{\log(1-t)}{t}\, dt \)

\(\displaystyle I=\int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt -\log(2) \text{Li}_2(1) = \int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t}- \frac{\pi^2}{6} \log(2)\)

The next step might be to evaluate

\(\displaystyle \int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt \)
 
  • #6
ZaidAlyafey said:
Here is an integral that I seen in another forum

\(\displaystyle \int^1_0 \frac{\log(x)\log(x+1)}{1-x} \approx -0.5 \)

That was a value return by W|A . This thread will be dedicated to try to solve the integral . Any ideas will always be welcomed .

The integral might require some dilogarithm identities . The first step might be integration by parts .

Integrals of this type are analized in...

http://www.mathhelpboards.com/f49/integrals-natural-logarithm-5286/#post24247

... and in particular for an analytic function...

$$f(x) = \sum_{k=0}^{\infty} a_{k}\ x^{k}\ (1)$$... the following formula holds...

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

In Your case is $f(x)= \frac{\ln (1+x)}{1-x}$ and is...

$$\frac{\ln (1+x)}{1-x} = x + \frac{x^{2}}{2} + \frac{5\ x^{3}}{6} + \frac{7\ x^{4}}{12} + \frac{47\ x^{5}}{60} + \frac{37\ x^{6}}{60} + \frac{319\ x^{7}}{420} + \frac{533\ x^{8}}{540} + ...\ (3)$$

... so that is...

$$\int_{0}^{1} \frac{\ln (1+x)}{1-x}\ \ln x\ dx = - \frac{1}{4} - \frac{1}{18} - \frac{5}{96} - \frac{7}{300} - \frac{47}{2160} - \frac{37}{2940} - \frac{319}{26680} - \frac{533}{43740} - ...\ (4)$$

Kind regards

$\chi$ $\sigma$
 
  • #7
In the previous post we got the following integral

\(\displaystyle \int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt \)

After integration by parts we get

\(\displaystyle \int^1_0 \frac{\log(1-t)\log(1-\frac{t}{2})}{t} \, dt = \frac{\pi^2}{6} \log(2) -\int^1_0 \frac{\text{Li}_2(x)}{2-x}\, dx\)

Adding the results together we get \(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x}\, dx =\int^1_0 \frac{\text{Li}_2(x)}{x-2}\, dx \)
 
  • #8
\(\displaystyle \int^1_0 \frac{\text{Li}_2(x)}{x-2}\, dx \)

\(\displaystyle \frac{-1}{2} \int^1_0 \frac{\text{Li}_2(x)}{1-\frac{x}{2}}\, dx \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0} \int^1_0 x^k \text{Li}_2(x)\, dx \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0} \sum_{n\geq 1}\frac{1}{2^k n^2} \int^1_0 x^{k+n} dx \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0} \sum_{n\geq 1}\frac{1}{2^k}\frac{1}{n^2(n+k+1)} \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0}\frac{1}{2^k} \sum_{n\geq 1}\frac{1}{n^2(k+1)}+\frac{(k+1)}{n(k+1)^2(n+k+1)} \)

\(\displaystyle \frac{-1}{2}\sum_{k\geq 0}\frac{1}{2^k} \sum_{n\geq 1}\frac{1}{n^2(k+1)}\,\,+\,\,\frac{1}{2}\sum_{k\geq 0}\frac{1}{2^k} \sum_{n\geq 1}\frac{(k+1)}{n(k+1)^2(n+k+1)} \)

\(\displaystyle \frac{-\zeta(2)}{2}\sum_{k\geq 0}\frac{1}{2^k(k+1)}\,\, +\,\, \frac{1}{2}\sum_{k\geq 0}\frac{1}{2^k(k+1)^2 } \sum_{n\geq 1}\frac{(k+1)}{n(n+k+1)} \)

\(\displaystyle -\zeta(2) \log(2) +\sum_{k\geq 0}\frac{\psi(k+2) + \gamma}{2^{k+1}(k+1)^2 } \)

\(\displaystyle -\zeta(2) \log(2) +\sum_{k\geq 0}\frac{H_{k+1}}{2^{k+1}(k+1)^2 } \)

\(\displaystyle -\zeta(2) \log(2) +\sum_{k\geq 1}\frac{H_{k}}{2^{k}k^2 } \)

Last step is solve the Euler sum

\(\displaystyle \sum_{k\geq 1}\frac{H_{k}}{2^{k}k^2 } \)
 
  • #9
In the previous post I used the following relation

\(\displaystyle \psi(n) = H_{n-1}-\gamma\)

Proof

By definition we have

\(\displaystyle \psi(n) = -\gamma +\sum_{k\geq 0}\frac{1}{k+1}-\frac{1}{k+n} \)

Since $n$ is an integer we can do the following

\(\displaystyle \sum_{k\geq 0}\frac{1}{k+n} = \sum_{k \geq n}\frac{1}{k} \)

\(\displaystyle \psi(n) = -\gamma +\sum_{k\geq 1}\frac{1}{k}- \sum_{k \geq n}\frac{1}{k} = -\gamma + \sum_{k=1}^{n-1}\frac{1}{k} \)

Now since harmonic numbers are defined as follows

\(\displaystyle \sum_{k=1}^{n}\frac{1}{k} = H_n\)

The proof is complete \(\displaystyle \square \).
 
  • #10
An attempt to solve the Euler sum

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2}\)

Rather we will try to generalize a little bit

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{ k^2} x^k \)

Start by the following

\(\displaystyle \sum_{k\geq 1} H_k \, x^k = -\frac{\log(1-x)}{1-x}\)

\(\displaystyle \sum_{k\geq 1} H_k \, x^{k-1} = -\frac{\log(1-x)}{x(1-x)} = - \frac{\log(1-x)}{x}- \frac{\log(1-x)}{1-x}\)

Integrating both sides we have

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{k} \, x^{k} = \text{Li}_2(x)+\frac{\log^2(1-x)}{2}\)

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{k} \, x^{k-1} = \frac{\text{Li}_2(x)}{x}+\frac{1}{2}\frac{\log^2(1-x)}{x}\)

Integrating again we obtain

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)+\frac{1}{2} \int^x_0 \frac{\log^2(1-t)}{t}\, dt\)

A further attempt will be in the next post .
 
  • #11
\(\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt \)

Integrating by parts we get the following

\(\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) -\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt \)

Now we are left with the following integral

\(\displaystyle \int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt = \int^{1}_{1-x} \frac{\text{Li}_2 (1-t)}{t} \, dt \)

\(\displaystyle \int^{1}_{1-x} \frac{\frac{\pi^2}{6} -\text{Li}_2(t) - \log(1-t) \log(t)}{t} \, dt \)

\(\displaystyle -\frac{\pi^2}{6}\log(1-x)- \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt -\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} \, dt \)

The first integral

  • \(\displaystyle \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt =\text{Li}_3(1)-\text{Li}_3(1-x) \)

The second integral by parts we obtain

  • \(\displaystyle \int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} =\text{Li}_3(1) +\log(1-x)\text{Li}_2(1-x)-\text{Li}_3(1-x) \)
Collecting the results together we obtain \(\displaystyle \int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt =-\frac{\pi^2}{6}\log(1-x)-\text{Li}_2(1-x) \log(1-x)+2\, \text{Li}_3(1-x)- 2 \zeta(3) \)Hence we solved the integral \(\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) \)

So we have got our Harmonic sum \(\displaystyle \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)+\frac{1}{2} \left( - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) \right) \)\(\displaystyle \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3) \)The expression can be further simplified but I will leave it like this
So we finally we got our solution (Whew)
 
  • #12
ZaidAlyafey said:
It is a based e logarithm . In complex analysis we always assume \(\displaystyle \log\) to be the natural logarithm , it looks better than \(\displaystyle \ln \,\, \) (Cool).

That's debatable. I think it looks better as ln, and it is clearly less time consuming to write ln :P
 
  • #13
We already proved that

\(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x} \, dx = -\zeta(2) \log(2) +\sum_{k\geq 1}\frac{H_{k}}{2^{k}k^2 }\)

We found that

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)\)

putting \(\displaystyle x =1/2 \) we get

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2} = -\log(2) \text{Li}_2\left( \frac{1}{2}\right) -\frac{1}{2}\log^3(2)+\zeta(3)\)

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2} = -\log(2) \left( \frac{\pi^2}{12}- \frac{1}{2}\log^2(2) \right) -\frac{1}{2}\log^3(2)+\zeta(3)\)

\(\displaystyle \sum_{k\geq 1} \frac{H_k}{2^k k^2} = -\frac{\pi^2}{12} \log(2)+\zeta(3)\)

\(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x} \, dx = -\frac{\pi^2}{6}\log(2) -\frac{\pi^2}{12} \log(2)+\zeta(3)\)

Hence we found the result was seeking for

\(\displaystyle \int^1_0 \frac{\log(x) \log(1+x)}{1-x} \, dx = -\frac{\pi^2}{4}\log(2)+\zeta(3) \)​
 

FAQ: An interesting logarithm integral

What is a logarithm integral?

A logarithm integral is a mathematical concept that involves integrating a logarithmic function. The most common form of a logarithm integral is the natural logarithm, which is written as ln(x). This integral is used to calculate the area under the curve of a logarithmic function.

How is a logarithm integral solved?

A logarithm integral is solved using various techniques such as integration by parts, substitution, or the use of logarithmic identities. It is important to simplify the integral as much as possible before attempting to solve it using one of these techniques.

What are the applications of a logarithm integral?

A logarithm integral has numerous applications in mathematics, physics, engineering, and economics. It is commonly used to model exponential growth and decay, calculate probabilities in statistics, and solve differential equations.

What is the difference between a logarithm integral and a regular integral?

A logarithm integral involves integrating a logarithmic function, while a regular integral involves integrating any other type of function. The techniques used to solve these integrals may differ, and the resulting answers may also be different.

Are there any special properties of logarithm integrals?

Yes, there are several special properties of logarithm integrals. For example, the integral of ln(x) is equal to xln(x) - x + C, where C is a constant. Additionally, the integral of a logarithmic function multiplied by a constant is equal to the constant multiplied by the integral of the logarithmic function.

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