How Does One Solve the Integral of Logarithmic Functions Over a Unit Interval?

In summary, a logarithm integral is a mathematical function denoted by "Li(x)" that calculates the area under the curve of the natural logarithm function. It can be evaluated using numerical methods or approximated with series expansions or special functions. It has various applications in mathematics, physics, and engineering. While there are methods for evaluating it, there is no closed-form solution for all values of x and there are open questions and conjectures related to it. The logarithm integral is a non-elementary function with interesting connections to other mathematical concepts.
  • #1
alyafey22
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logarithm integral # (2)

Solve the following

\(\displaystyle \int^1_0 \frac{\log(1+x)\log(1-x)}{x}\,dx\)​
 
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  • #2
ZaidAlyafey said:
Solve the following

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

$$\log (1-x)=-\sum_{n=1}^{\infty} \frac{x^n}{n} \text{ for } |x|<1$$

$$\log (1+x)=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \text{ for } |x|<1$$

So:

$$\frac{\log(1-x)\log(1+x)}{x}=\frac{- \sum_{n=1}^{\infty} \frac{x^n}{n} \cdot \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}}{x}=\frac{\sum_{n=1}^{\infty} (-1)^{n+2} \frac{x^{2n}}{n^2}}{x}=\sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{n^2}$$

Therefore:

$$\int_0^1 \frac{\log(1+x) \log(1-x)}{x} dx=\int_0^1 \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{n^2} dx=\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \left [ \frac{x^{2n}}{2n}\right ]_0^1=\frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$$
 
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  • #3
evinda said:
$$\frac{\log(1-x)\log(1+x)}{x}=\frac{- \sum_{n=1}^{\infty} \frac{x^n}{n} \cdot \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}}{x}=\frac{\sum_{n=1}^{\infty} (-1)^{n+2} \frac{x^{2n}}{n^2}}{x}=\sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{n^2}$$

How did you perform the multiplication of the two series ?
 
  • #4
In evinda's solution, there is an error made in multiplying the two series. The product of two absolutely convergent power series is given by the Cauchy product, not pointwise product.
 
  • #5
Euge said:
In evinda's solution, there is an error made in multiplying the two series. The product of two absolutely convergent power series is given by the Cauchy product, not pointwise product.

I see..I am sorry! (Tmi)
 
  • #6
$$\int^1_0 \frac{\log(1-x)\log(1+x)}{x}\,dx = -\zeta(2)\log(2)+\int^1_0 \frac{\mathrm{Li}_2(x)}{1+x}\,dx $$

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1+x}\,dx = \int^1_0 \sum_k (-x)^k \mathrm{Li}_2(x) = \sum_{k\geq 1} (-1)^{k-1}\int^1_0 x^{k-1} \mathrm{Li}_2(x) $$

Since

$$\int^1_0 x^{k-1}\mathrm{Li}_2(x)\, dx = \sum_{n\geq 1}\frac{1}{n^2(n+k)} =\frac{1}{k}\left(\zeta(2)-\frac{H_k}{k}\right)$$

We've got

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1+x}\,dx =\zeta(2)\sum_{k\geq 1} \frac{(-1)^{k-1}}{k}-\sum_{k\geq 1} (-1)^{k-1}\frac{H_k}{k^2} = \frac{1}{24}\left(4 \pi^2 \log(2)-15 \zeta(3)\right)$$

$$\int^1_0 \frac{\log(1-x)\log(1+x)}{x}\,dx = -\zeta(2)\log(2)+\zeta(2)\log(2)- \frac{15}{24} \zeta(3)=\frac{5}{8}\zeta(3)$$
 

FAQ: How Does One Solve the Integral of Logarithmic Functions Over a Unit Interval?

1. What is a logarithm integral?

A logarithm integral is a mathematical function that calculates the area under the curve of the natural logarithm function. It is denoted by the symbol "Li(x)" and is defined as the integral from 0 to x of 1/ln(t) dt.

2. How is a logarithm integral evaluated?

A logarithm integral can be evaluated using numerical methods, such as Simpson's rule or the trapezoidal rule. It can also be approximated using series expansions or special functions, such as the polylogarithm function.

3. What are the applications of a logarithm integral?

A logarithm integral has various applications in mathematics, physics, and engineering. It is used to solve differential equations, analyze the distribution of primes, calculate complex integrals, and model phenomena such as heat transfer and diffusion.

4. Is the logarithm integral a solved problem?

No, the logarithm integral is not a completely solved problem. While there are various methods for evaluating it, there is no closed-form solution for all values of x. Additionally, there are many open questions and conjectures related to the logarithm integral.

5. Are there any special properties of the logarithm integral?

Yes, the logarithm integral has several interesting properties, such as the fact that it is a non-elementary function and cannot be expressed in terms of elementary functions. It also has connections to other mathematical concepts, such as the Riemann zeta function and the prime number theorem.

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