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} $.

4 comments:

  1. the inside sum telescopes, the outside sum is evaluated by using the asymptotic form of the harmonic series

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  2. the solution that only involves partial fractions for the practice problem turns out to be quite ugly.

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  3. looking at your integral, reminds me of the following quite nice problem:

    evaluate the int_0^1 ln(1-x) ln(1+x) dx.

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  4. Yeah, I saw that on AoPS. The solution was a series expansion though... and then using the harmonic series, right?

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