A tricky logarithmic integral

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \frac{1-x}{\log x} \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x\]

Solution

Well, we all have the basic idea that we are going to kill the logarithm by inserting a parameter , call it \alpha , and then differentiate under the integral sign but doing so we have that if we define

    \[f\left ( \alpha \right ) = \int_{0}^{1} \frac{1-x^\alpha}{\log x} \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \quad , \quad \alpha \geq 0\]

then

    \begin{align*} \frac{\mathrm{d} }{\mathrm{d} \alpha} f(\alpha) &= \int_{0}^{1} \frac{\partial }{\partial \alpha} \frac{1-x^\alpha}{\log x} \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \\ &=-\int_{0}^{1} x^\alpha \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} x^{2^n} x^\alpha \, {\rm d}x\\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} x^{2^n +\alpha} \, {\rm d}x \\ &=- \sum_{n=0}^{\infty} \frac{1}{\alpha +2^n +1} \end{align*}

And this is why the integral is tricky. You cannot find a general closed form for the derivative but that doesn’t matter because all you care about is to evaluate f(1). Since f(0)=0 we have that

    \[f(1)=\int_0^1 f'(\alpha) \, {\rm d}\alpha\]

Therefore,

    \begin{align*} f(1) &= \int_{0}^{1} f'(\alpha) \, {\rm d}\alpha \\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} \frac{{\rm d}\alpha}{\alpha +2^n +1}\\ &= -\sum_{n=0}^{\infty} \log \left ( \frac{2^n +2}{2^n +1} \right )\\ &= - \lim_{N \rightarrow +\infty} \sum_{n=0}^{N} \log \left ( \frac{2^n +2}{2^n +1} \right ) \\ &= - \lim_{N \rightarrow +\infty} \log \left ( \frac{3 \cdot 2^N}{2^N +1} \right ) \\ &= - \log 3 \end{align*}

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