# Digamma fourth integral representation proof

$$\psi(z) = \log(z) -\frac{1}{2z}-2\int^\infty_0 \frac{t}{(t^2+z^2)(e^{2\pi}-1)}dt\,\,\,\,;\text{ Re}z>0$$

We prove that

$$2\int^\infty_0 \frac{t}{(t^2+z^2)(e^{2\pi}-1)}dt= \log(z) -\frac{1}{2z}- \psi(z)$$
First note that

$$\frac{2}{e^{2\pi t}-1} =\coth(\pi t)-1$$

Also note that

$$\coth(\pi t) = \frac{1}{\pi t}+\frac{2t}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2}$$

Hence we conclude that

$$\frac{2t}{e^{2\pi t}-1} =\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2}$$

Substitute the value in the integral

$$\int^\infty_0 \frac{1}{t^2+z^2}\left\{\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2} \right\}dt$$

The first integral

$$\frac{1}{\pi}\int^\infty_0 \frac{1}{t^2+z^2}dt = \frac{1}{2z}$$

Since the second integral is divergent we put

$$\int^N_0 \frac{t}{t^2+z^2}\,dt = \frac{1}{2}\log(N^2+z^2)-\log(z)$$

Also for the series

$$\frac{2}{\pi}\sum_{k=1}^N\int^\infty_0 \frac{t^2}{(t^2+z^2)(t^2+k^2)}dt =\sum_{k=1}^N\frac{1}{k+z}$$

Which simplifies to

$$\sum_{k=1}^N\frac{1}{k+z} = \sum_{k=1}^N\frac{1}{k}-\sum_{k=1}^N\frac{z}{k(k+z)} = H_N – \sum_{k=1}^N\frac{z}{k(k+z)}$$

Now take the limit

$$\lim_{N \to \infty} -\frac{1}{2}\log(N^2+z^2)+\log(z) +H_N – \sum_{k=1}^N\frac{z}{k(k+z)}$$

Or

$$\lim_{N \to \infty} H_N-\log(N)+\log(z)- \sum_{k=1}^\infty\frac{z}{k(k+z)}$$

This simplifies to

$$\log(z) +\gamma- \sum_{k=1}^\infty\frac{z}{k(k+z)} = \log(z) -\frac{1}{z}-\psi(z)$$

Collecting the results we have

$$2\int^\infty_0 \frac{t}{(t^2+z^2)(e^{2\pi}-1)}dt = \log(z)+ \frac{1}{2z}-\frac{1}{z}-\psi(z) = \log(z)- \frac{1}{2z}-\psi(z)$$

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