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Integral representation of gamma

Let \gamma denote the Euler – Mascheroni constant. Define F(x)=\sum \limits_{n=1}^{\infty} x^{2^n}. Prove that

    \[\gamma = 1 - \int_{0}^{1} \frac{F(x)}{1+x} \, {\rm d}x\]

Solution

Well let F_n(x)=\sum \limits_{k=1}^n x^{2^k}; since F_n increase to F on [0, 1) we deduce by the monotone convergence theorem that

    \[\lim_{n\to +\infty}\int_0^1 \frac{F_n(x)}{1+x}\,{\rm d} x=\int_0^1 \frac{F(x)}{1+x}\,{\rm d}x\]

Now,

     \begin{align*} \int_0^1 \frac{F_n(x)}{1+x}\,{\rm d}x &= \sum_{k=1}^n \left[\int_0^1 \frac{x^{2^k}}{1+x}\,{\rm d}x\right]\\ &= \sum_{k=1}^n \left[\int_0^1 \frac1{1+x}\,{\rm d} x - \int_0^1 \frac{1-x^{2^k}}{1+x}\,{\rm d}x\right]\\ &= \sum_{k=1}^n \bigg[\ln 2 - \int_0^1 \bigg(1-x+x^2- \\ & \quad \quad  - \cdots-x^{2^k-1} \bigg) \,{\rm d}x \bigg]\\ &= \sum_{k=1}^n \left[\ln 2 -\sum_{m=1}^{2^k}\frac{(-1)^{m-1}}{m}\right]\\ &= \sum_{k=1}^n \left[\ln 2 -\sum_{j=2^{k-1}+1}^{2^k}\frac1j\right]\\ &= n\ln 2 -\sum_{j=2}^{2^n}\frac1j \\ &= 1 + \ln 2^n - \sum_{j=1}^{2^n} \frac{1}{j} \end{align*}

Hence

    \begin{align*} \int_{0}^{1} \frac{F(x)}{1+x} \, {\rm d}x &= \int_{0}^{1} \frac{1}{1+x} \lim_{n \rightarrow +\infty} F_n(x) \, {\rm d}x \\ &=\lim_{n \rightarrow +\infty} \int_{0}^{1} \frac{F_n(x)}{1+x} \, {\rm d}x \\ &= \lim_{n \rightarrow +\infty} \left [ 1 + \ln 2^n - \sum_{j=1}^{2^n} \frac{1}{j} \right ]\\ &= 1 - \gamma \end{align*}

 

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1 Comment

  1. The above identity is due to Catalan (1875). The F above also satisfies the relation

    (1)   \begin{equation*} F(x) = x^2 + F \left( x^2 \right) \end{equation*}

    which is a typical example of Mahler’s functional equations in the theory of transcedental numbers. Also note that

    (2)   \begin{equation*} \int_0^1 \frac{F(x)}{x} \, {\rm d}x = \sum_{n=1}^{\infty} \frac{1}{2^n}=1 \end{equation*}

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