## Friday, December 1, 2006

### Integra-what? Topic: Calculus/S&S.

Problem: Write $I = \displaystyle \sum_{l_1=1}^{\infty} \sum_{l_2=1}^{\infty} \frac{(-1)^{l_1-1}}{l_1l_2(l_1+l_2)}$ as an integral.

Solution: We first notice that

$\displaystyle \frac{(-1)^{l_1-1}}{l_1l_2(l_1+l_2)} = \frac{1}{l_1l_2} \int_0^1 (-x)^{l_1-1}x^{l_2} dx$.

So our summation becomes

$\displaystyle I = \sum_{l_1=1}^{\infty} \sum_{l_2=1}^{\infty} \frac{1}{l_1l_2} \int_0^1 (-x)^{l_1-1}x^{l_2} dx$.

But we can switch the order of the integral with the two sums because $\int (f+g) = \int f+\int g$. Hence we obtain

$\displaystyle I = \int_0^1 \sum_{l_1=1}^{\infty} \left(\frac{(-x)^{l_1-1}}{l_1}\sum_{l_2=1}^{\infty} \frac{x^{l_2}}{l_2} \right)$.

We remember, however, that the Taylor series expansion of $\displaystyle \ln{(1-x)} = -\sum_{n=1}^{\infty} \frac{x^n}{n}$, so substituting we have

$\displaystyle I = -\int_0^1 \ln{(1-x)}\sum_{l_1=1}^{\infty} \frac{(-x)^{l_1-1}}{l_1}$.

Similarly, $\displaystyle \ln{(1+x)} = -\sum_{n=1}^{\infty} \frac{(-x)^n}{n}$, so we replace one more time to get

$\displaystyle I = -\int_0^1 \frac{1}{x} \ln{(1-x)} \ln{(1+x)} dx$.

QED.

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Comment: Transforming sums into integrals and vice versa are powerful techniques for evaluating these expressions. The numerical value turns out to be $I = \frac{5}{8} \zeta(3)$, but we'll save that for another time...

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Practice Problem: (1978 Putnam - B2) Evaluate

$\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2n+n^2m+2mn}$.