First integral representation of digamma proof

$$ \psi(a) = \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\,dz $$

We begin with the double integral

$$\int^{\infty}_0 \int^t_1 \,e^{-xz}\,dx\,dz=\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz $$

Using fubini theorem we also have

$$ \int^t_1 \int^{\infty}_0\,e^{-xz}\,dz \,dx = \int^t_1 \frac{1}{x}\,dx = \log t

Hence we have the following

$$\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz = \log(t)$$

We also know that

$$\Gamma'(a) = \int^{\infty}_0 t^{a-1}e^{-t}\,\log t \, dt$$

Hence we have

$$\Gamma'(a) = \int^{\infty}_0 t^{a-1}e^{-t}\,\left(\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz \right)\, dt = \int^{\infty}_0 \int^{\infty}_{0}\frac{t^{a-1}e^{-t}e^{-z}-t^{a-1}e^{-t(z+1)}}{z}\, dz \, dt$$

Now we can use the fubini theorem

$$\Gamma'(a) = \int^{\infty}_0 \int^{\infty}_{0}\frac{t^{a-1}e^{-t}e^{-z}-t^{a-1}e^{-t(z+1)}}{z}\, dt \, dz$$

$$\Gamma'(a) = \int^{\infty}_0 \frac{1}{z} \left( e^{-z}\int^{\infty}_{0}t^{a-1}e^{-t}\, dt-\int^{\infty}_0 t^{a-1}e^{-t(z+1)}\, dt \right)\, dz$$

But we can easily deduce using Laplace that

$$\int^{\infty}_0 t^{a-1}e^{-t(z+1)}\,dt= \Gamma(a) \,(z+1)^{-a}

Aslo we have

$$\int^{\infty}_{0}t^{a-1}e^{-t}\, dt=\Gamma(a)

Hence we can simplify our integral to the following

$$\Gamma'(a) = \Gamma(a)\, \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z} dz$$

$$\frac{\Gamma'(a)}{\Gamma(a)} = \psi(a) = \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\, dz $$

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