**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.

--------------------

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...

--------------------

Practice Problem: (1978 Putnam - B2) Evaluate

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

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

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

ReplyDeletelooking at your integral, reminds me of the following quite nice problem:

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

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

ReplyDelete