Digamma third integral representation proof

$$\psi \left(a\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(a t\right)}}{1-e^{-t}}\, dt$$

$$\textit{proof}$$

Let \( e^{-t}=x \)

$$\int^{1}_0 \, -\frac{1}{\log(x)}-\frac{x^{a-1}}{1-x}\, dx$$

By adding and subtracting 1

$$-\int^{1}_0 \, \frac{1}{\log(x)}+\frac{1}{1-x}\, dx+\int^1_0\frac{1-x^{a-1}}{1-x}\, dx$$

Using the second integral representation

$$-\int^{1}_0 \, \frac{1}{\log(x)}+\frac{1}{1-x}\, dx+\gamma+\psi(a)$$

We can prove that

$$\int^{1}_0 \, \frac{1}{\log(x)}+\frac{1}{1-x}\, dx = \gamma$$

Finally we get

$$\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(a t\right)}}{1-e^{-t}}\, dt=-\gamma+\gamma+\psi(a) = \psi(a)$$

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