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please how we calculate $$-\displaystyle\int_{0}^1 \dfrac{\ln(x)}{x^2-1} dx$$ and $$\displaystyle\int_{1}^{+\infty} \dfrac{\ln(x)}{x^2-1} dx$$ ?

i try with integration by part we have $-\displaystyle\int_0^1 \dfrac{\ln (x)}{x^2-1} dx = -\lim_{a \to 0} \displaystyle\int_a^1 \dfrac{\ln (x)}{x^2-1}$

with $u(x)=\ln(x)$ then $u'(x)=\dfrac{1}{x}$ and $v'(x)=\dfrac{1}{x^2-1}$ but who is $v$? i have no result. thank you in advance to help me

user487908
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2 Answers2

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Well, we have:

$$\mathscr{I}_{\space\text{n}}:=\int_1^\text{n}\frac{\ln\left(x\right)}{x^2-1}\space\text{d}x=\frac{1}{2}\cdot\left\{\int_1^\text{n}\frac{\ln\left(x\right)}{x-1}\space\text{d}x-\int_1^\text{n}\frac{\ln\left(x\right)}{x+1}\space\text{d}x\right\}\tag1$$

Using integration by parts (IBP):

$$\int_1^\text{n}\frac{\ln\left(x\right)}{x+1}\space\text{d}x=\left[\ln\left(x\right)\cdot\ln\left(x+1\right)\right]_1^\text{n}-\int_1^\text{n}\frac{\ln\left(x+1\right)}{x}\space\text{d}x=$$ $$\ln\left(\text{n}\right)\cdot\ln\left(\text{n}+1\right)-\int_1^\text{n}\frac{\ln\left(x+1\right)}{x}\space\text{d}x\tag2$$

Let $\text{u}=-x$:

$$\int_1^\text{n}\frac{\ln\left(x+1\right)}{x}\space\text{d}x=-\int_{-1}^{-\text{n}}-\frac{\ln\left(1-\text{u}\right)}{\text{u}}\space\text{d}\text{u}=$$ $$-\left[\text{Li}_2\left(\text{u}\right)\right]_{-1}^{-\text{n}}=-\left(\frac{\pi^2}{12}+\text{Li}_2\left(-\text{n}\right)\right)\tag3$$

Where $\text{Li}_2\left(\text{u}\right)$ is the dilogarithm.

Substitute $\text{z}=x-1$:

$$\int_1^\text{n}\frac{\ln\left(x\right)}{x-1}\space\text{d}x=\int_0^{\text{n}-1}\frac{\ln\left(\text{z}+1\right)}{\text{z}}\space\text{d}\text{z}=$$ $$\left[-\text{Li}_2\left(-\text{z}\right)\right]_0^{\text{n}-1}=-\text{Li}_2\left(1-\text{n}\right)\tag4$$

So, we end up with:

$$\mathscr{I}_{\space\text{n}}=\frac{1}{2}\cdot\left\{\text{Li}_2\left(1-\text{n}\right)-\left(\frac{\pi^2}{12}+\text{Li}_2\left(-\text{n}\right)\right)\right\}\tag5$$

When $\text{n}\to\infty$:

$$\lim_{\text{n}\to\infty}\mathscr{I}_{\space\text{n}}=-\frac{\pi^2}{24}\tag6$$

Community
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Jan Eerland
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  • Please, i don't understand how we calculate $-\displaystyle\int_{1}^n - \dfrac{\ln(1-u)}{u} du$ – user487908 Nov 23 '17 at 19:24
  • Look at the link that I provided in my answer. – Jan Eerland Nov 23 '17 at 19:56
  • but i dont undestrand what is $Li_2(u)$, i never work with it – user487908 Nov 23 '17 at 20:22
  • @user487908 That's why I provided that link. – Jan Eerland Nov 23 '17 at 20:28
  • Please, but for $\displaystyle\int_0^1 \dfrac{\ln(x)}{x^2-1}$ we have $=\dfrac{1}{2} {\displaystyle\int_0^1 \dfrac{\ln x}{x-1} dx - \displaystyle\int_0^1 \dfrac{\ln x}{x+1} dx$. How we calculate these two integrals using $L_i$? Please – user487908 Nov 23 '17 at 22:57
  • i try to do this: we put $x=1-z$ then $$\displaystyle\int_0^1 \dfrac{\ln(x)}{x-1}dx=\displaystyle\int_1^0 \dfrac{\ln(1-z)}{-z} dz= Li_2(1)$$ how we calculate $Li_2(1)$? and for $\displaystyle\int_0^1 \dfrac{\ln(x)}{x+1}dx$? – user487908 Nov 23 '17 at 23:27
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Here is an approach for evaluating your two integrals that does not make use of the dilogarithmic function. We will instead make use of the following well-known result of $$\sum^\infty_{n = 1} \frac{1}{n^2} = \frac{\pi^2}{6},$$ which is known as the Basel problem.

First consider the integral $$I = \int^\infty_1 \frac{\ln x}{x^2 - 1} \, dx.$$ Setting $x = 1/u, dx = -1/u^2 \, du$ while for the limits of integration we have $(1,\infty) \mapsto (1,0)$. Thus $$I = \int^1_0 \frac{\ln \left (\frac{1}{u} \right )}{\frac{1}{u^2} - 1} \frac{du}{u^2} = \int^1_0 \frac{-\ln u}{1 - u^2} \,du = \int^1_0 \frac{\ln u}{u^2 - 1}.$$ So we see $$\int^\infty_1 \frac{\ln x}{x^2 - 1} \, dx = \int^1_0 \frac{\ln x}{x^2 - 1} \, dx.$$

Now consider the right most integral. Recognising the term $\dfrac{1}{1 - x^2}$ as the sum of a convergent geometric series, that is $$\frac{1}{x^2 - 1} = - \frac{1}{1 - x^2} = -\sum^\infty_{n = 0} x^{2n}, \quad |x| < 1$$ the integral can be rewritten as $$I = -\int^1_0 \ln x \sum^\infty_{n = 0} x^{2n} \, dx.$$ Interchanging the integral sign with the summation (see here for when one is exactly allowed to do this) gives $$I = -\sum^\infty_{n = 0} \int^1_0 x^{2n} \ln x \, dx.$$

Integrating by parts gives $$I = \sum^\infty_{n = 0} \frac{1}{2n + 1} \int^1_0 x^{2n} \, dx,$$ while integrating again gives $$I = \sum^\infty_{n = 0} \frac{1}{(2n + 1)^2}.$$

The sum appearing above is well known and can be reduce to the Basel problem as follows: \begin{align*} \sum^\infty_{n = 0} \frac{1}{(2n + 1)^2} &= \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots\\ &= \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots \right ) - \left (\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots \right )\\ &= \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots \right ) - \frac{1}{2^2} \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots \right )\\ &= \sum^\infty_{n = 1} \frac{1}{n^2} - \frac{1}{2^2} \sum^\infty_{n = 1} \frac{1}{n^2}\\ &= \frac{3}{4} \sum^\infty_{n = 1} \frac{1}{n^2}\\ &= \frac{3}{4} \cdot \frac{\pi^2}{6}\\ &= \frac{\pi^2}{8}. \end{align*}

Thus $$\int^\infty_1 \frac{\ln x}{x^2 - 1} \, dx = \int^1_0 \frac{\ln x}{x^2 - 1} \, dx = \frac{\pi^2}{8}.$$

omegadot
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  • Thank youso much for this answer. There is just an problrm. We have $$\displaystyle\int_1^{+\infty} \dfrac{\ln(x)}{x^2-1}dx= \displaystyle\int_0^1\dfrac{\ln(u)}{1-u^2} du$ beacause $dx=-\dfrac{1}{u^2} du$. You lost "-" in the entegral. No? – user487908 Nov 24 '17 at 09:13
  • No because that negative sign was used in reversing the order of the limits of integration. – omegadot Nov 24 '17 at 09:18
  • Ah ok! Thank you so much. Then the final result $$\displaystyle\int_0^{+\infty} \dfrac{\ln x}{x^2-1} dx= \displaystyle\int_0^1\dfrac{\ln x}{x^2-1} dx+ \displaystyle\int_1^{+\infty}\dfrac{\ln x}{x^2-1} dx= \dfrac{\pi^2}{4}$? And why we have to decouplate the integral $\displaystyle\int_0^{+\infty} |\dfrac{\ln (x)}{x^2-1}|dx$ to calculate it? Please – user487908 Nov 24 '17 at 09:25
  • Yes, so adding the two integrals I evaluated together will indeed give the result of $\pi^2/4$ for the integral from zero to $\infty$. There are many different ways; in fact a bewildering number of ways; in which improper integrals may be evaluated. Dividing the interval $(0,\infty)$ into two intervals, one of $(0,1)$ the other of $(1,\infty)$, is a particularly common trick one can use particularly when you have a logarithmic term appearing in the numerator. Of course such a method does not always work and one only gets good at recognising when this can be done through practice. – omegadot Nov 24 '17 at 11:45
  • Thak you so much for the help. – user487908 Nov 24 '17 at 12:21