Evaluate: $$\int^0_1 \dfrac{\ln(t)}{1-t^2}dt$$

This actually came up while solving another integral. It was suggested that I use a binomial series, but unfortunately I do not understand how to use this. Can anyone help me out?

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    $\begingroup$ $$\sum_{k=0}^\infty \int_0^1 t^{2k}(-\ln t)\,dt,$$ then you can for example substitute $t = e^{-u}$. You get a series representation of the value of the integral. I think the series is not unknown. $\endgroup$ – Daniel Fischer Jun 22 '15 at 11:42
  • $\begingroup$ Sorry Sir, but I couldn't understand what you did. Sir, how did you transform the integral into $$\sum_{k=0}^\infty \int_0^1 t^{2k}(-\ln t)\,dt?$$ $\endgroup$ – User1234 Jun 22 '15 at 11:44
  • $\begingroup$ Hint: $~\displaystyle\sum_{n=0}^\infty u^n~=~\dfrac1{1-u}~$ for $~|u|<1$. $\endgroup$ – Lucian Jun 22 '15 at 11:44
  • $\begingroup$ see this: math.stackexchange.com/questions/1334561/… $\endgroup$ – Math-fun Jun 22 '15 at 12:03
  • $\begingroup$ How can the 0 be on top and the 1 below? Is this a mistake or just something I haven't learned yet? I am in doubt because people also use it in the answer. $\endgroup$ – wythagoras Jun 22 '15 at 13:32

First consider the operation \begin{align} \partial_{n} \, t^{n} = \frac{d}{dn} \, e^{n \ln(t)} = \ln(t) \, e^{n \ln(t)} = t^{n} \, \ln(t). \end{align} Now consider the integral, where the operation just presented will be used, \begin{align} I_{n} = \int_{0}^{1} \ln(t) \, t^{n} \, dt = \partial_{n} \, \int_{0}^{1} t^{n} \, dt = \partial_{n} \left[ \frac{t^{n+1}}{n+1} \right]_{0}^{1} = \partial_{n} \left(\frac{1}{n+1}\right) = - \frac{1}{(n+1)^{2}} \end{align} Now letting $n \to 2n$ and then summing over $n$ it is seen that: \begin{align} \sum_{n=0}^{\infty} I_{2n} = \int_{0}^{1} \frac{\ln(t) \, dt}{1-t^{2}} &= - \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} \\ &= - \left(\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} \right) + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \\ &= - \sum_{n=1}^{\infty} \frac{1}{n^{2}} + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}} = - \zeta(2) + \frac{1}{4} \, \zeta(2) \\ &= - \frac{3}{4} \zeta(2) = - \frac{\pi^{2}}{8}. \end{align} The integral desired is: \begin{align} \int_{0}^{1} \frac{\ln(t) \, dt}{1-t^{2}} = - \frac{\pi^{2}}{8}. \end{align}

  • $\begingroup$ You might elaborate that $\partial_n$ means $\frac{\partial}{\partial n}$ and give some justification for the operations, in particular the interchange operations. $\endgroup$ – Ian Jun 22 '15 at 13:36
  • $\begingroup$ Sir, does it help in noticing that $I_n=\dfrac{\partial}{\partial n} \beta (n,y)\bigg|_{y=0}?$$$$$Sir, unfortunately I couldn't understand why $n\to 2n$ as well as why we took the sum over $n$. Sir, could you also explain how $$- \left(\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} \right) + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}} $$ $$= - \sum_{n=1}^{\infty} \frac{1}{n^{2}} + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}} ?$$ $\endgroup$ – User1234 Jun 22 '15 at 13:53
  • $\begingroup$ @BetterWorld First Leucippus did the integral calculation for an arbitrary $n$. Then they substituted in $2n$, since your quantity only involves $2n$. Plugging in you get that sum of the reciprocal of the squares of the odd natural numbers. We know the sum of the reciprocal of the squares of all the natural numbers from elsewhere. So Leucippus adds and subtracts the sum of the squares of the even natural numbers. Where it was added, now we know what to do. Where it was subtracted, you have the simple fact that $\sum_{n=1}^\infty\frac{1}{(2n)^2} = \frac{1}{4} \sum_{n=0}^\infty \frac{1}{n^2}$. $\endgroup$ – Ian Jun 22 '15 at 14:29
  • $\begingroup$ @BetterWorld Sorry, my very last sum should start at $1$, not $0$. $\endgroup$ – Ian Jun 22 '15 at 14:58
  • $\begingroup$ But Sir, shouldn't it be $$ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} =\sum_{n=1}^{\infty} \dfrac{1}{n^2} - \sum_{n=1}^{\infty}\dfrac{1}{(2n)^2}$$ $$\Rightarrow-\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} =-\bigg(\sum_{n=1}^{\infty} \dfrac{1}{n^2} - \sum_{n=1}^{\infty}\dfrac{1}{(2n)^2}\bigg)$$ $$\Longrightarrow -\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = -\bigg(\zeta(2)-\dfrac{1}{4}\zeta(2)\bigg)$$ $\endgroup$ – User1234 Jun 22 '15 at 15:28

The geometric series formula tells you that

$$\frac{1}{1-r}=\sum_{n=0}^\infty r^n$$

if $|r|<1$. Applying this to $r=t^2$ you get that your integral is

$$\int_1^0 \sum_{n=0}^\infty \ln(t) t^{2n} dt.$$

You can interchange the sum and integral, for example using monotone convergence (since the integrands are all negative), so you have

$$\sum_{n=0}^\infty \int_1^0 \ln(t) t^{2n} dt.$$

Each of these integrals can be done using integration by parts with $u=\ln(t)$ and $dv=t^{2n} dt$. They are improper at the endpoint of $0$, but this is no real obstacle, because the log term in each antiderivative is getting multiplied with a monomial, so the log terms in the definite integrals all vanish.

  • $\begingroup$ Sir, I understood the following: $$\dfrac{1}{1-t^2}=\sum_{n=0}^{n=\infty}t^{2n}$$ $$\Rightarrow\int_1^0 \dfrac{log t}{1-t^2}dt=\int_1^0log(t)\sum_{n=0}^{n=\infty}t^{2n}$$ $$=\int_1^0 \sum_{n=0}^{n=\infty}t^{2n}log(t)dt$$ But Sir, I couldn't understand how you changed the order of Summation and Integration. $\endgroup$ – User1234 Jun 22 '15 at 11:53
  • $\begingroup$ @BetterWorld Since $\ln(t) \leq 0$ and $t^{2n} \geq 0$, the product is negative, so $f_N(t)=\sum_{n=0}^N \ln(t)) t^{2n}$ is a decreasing sequence of functions. So you can use the monotone convergence theorem to get the result. Other options are available; for instance the convergence is uniform on $[\delta,1-\delta]$ for any $\delta>0$, so you might be able to use that, too. $\endgroup$ – Ian Jun 22 '15 at 12:07
  • $\begingroup$ Sir, I'm sorry but I couldn't understand the $f_N(t)$ notation. Also Sir, in general, when can we interchange the order of Summation and Integration? $\endgroup$ – User1234 Jun 22 '15 at 12:12
  • $\begingroup$ @BetterWorld That is a big question, which is one of the main topics in real analysis. The most general theorem I know of is the Vitali convergence theorem, although it does not help when the limiting integral is infinite. The most useful theorem I know of is the dominated convergence theorem, which can be regarded as a special case of the Vitali convergence theorem with convenient hypotheses. $\endgroup$ – Ian Jun 22 '15 at 12:21
  • $\begingroup$ Sir, over here it is said that if $f_n(x) \ge 0,$ $$\sum \int f_n(x) dx = \int \sum f_n(x) dx$$ Also, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$, then $$\int \sum f_n = \sum \int f_n$$ Sir, could you please help me understand the meaning of $f_n(x)?$ Could you also please tell me how to evaluate $ \int |f_n|?$ I've never Integrated the $modulus$ of a function before. $\endgroup$ – User1234 Jun 22 '15 at 12:37

Substitute $t\mapsto e^{-t}$: $$ \begin{align} \int_1^0\frac{\log(t)}{1-t^2}\,\mathrm{d}t &=\int_0^\infty\frac{t}{1-e^{-2t}}e^{-t}\,\mathrm{d}t\\ &=\sum_{k=0}^\infty\int_0^\infty te^{-(2k+1)t}\,\mathrm{d}t\\ &=\Gamma(2)\sum_{k=0}^\infty\frac1{(2k+1)^2}\\ &=\frac{\pi^2}8 \end{align} $$


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