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I have this problem from my calc II class. The question asks to express as an infinite sum $$ \int^1_0 \frac{\ln(x)}{1-x^2}\,dx $$ So far I've tried to rewrite the denominator as a power series and then multiply it by the numerator which gives $$ \int^1_0 \sum_{k=0}^{\infty}\ x^{2k}\ln(x)$$ From here I did integration by parts which is where I am unsure about how to proceed. What I've done was consider the integral in the sum as an improper integral and take the limit as $x$ approaches zero. The value of this limit is zero. Is it enough at this point to say that the value of the infinite sum is zero?

Bob Cobb
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We want $I_m = \displaystyle\int_0^1 x^m \ln(x) dx$. By integration by parts, we get that \begin{align} I_m & = \int_0^1 \ln(x) d\left(\dfrac{x^{m+1}}{m+1}\right) = \left.\dfrac{x^{m+1}}{m+1} \cdot \ln(x) \right \vert_0^1 - \int_0^1 \dfrac{x^{m+1}}{m+1} \dfrac{dx}x\\ & = - \int_0^1 \dfrac{x^m}{m+1}dx = - \dfrac1{(m+1)^2} \end{align} Now you should be able to finish it off. Move your cursor over the gray area for the complete answer. (The value of the infinite sum is not zero. The integrand is negative in the interval $[0,1]$.)

The integral you want to evaluate becomes (after swapping summation and limit)$$I_m = - \sum_{k=0}^{\infty} \dfrac1{(2k+1)^2} = \sum_{k=1}^{\infty} \dfrac1{(2k)^2} - \sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac14 \zeta(2) - \zeta(2) = - \dfrac34 \zeta(2) = - \dfrac{\pi^2}8$$

  • Why we can say $\int \sum = \sum \int$ ? – Cortizol Apr 20 '13 at 15:45
  • @Cortizol The sequence of partial sums form a monotone sequence to $\dfrac{\log(x)}{1-x^2}$. Hence, if the initial integral were to exist, and if the integral of the partial sums exist, then both the limit of the integral of the partial sums must equal the initial integral. –  Apr 20 '13 at 15:57