Derivative of the Gamma Function

Show that \({\Gamma}^{'} (1) = -\gamma \) where \(\gamma\) is the Euler-Mascheroni constant.

Solution

We begin with the integral definition of the Gamma function $\lim _{ x\rightarrow \infty }{ \int _{ 0 }^{ x }{ { e }^{ -t }{ t }^{ n-1 } } dt } .$

Let $f(n,t) = {e}^{-t}{t}^{n-1}$ and $\frac{\partial f}{\partial n} = {e}^{-t}{t}^{n-1} ln\left(t\right).$

By the Leibniz rule, $\lim _{ x\rightarrow \infty }{ \left[ \int _{ 0 }^{ x }{ { e }^{ -t }{ t }^{ n-1 }ln(t) } dt-{ e }^{ -x }{ x }^{ n-1 }-0 \right] }$ which reduces to $\int _{ 0 }^{ \infty }{ { e }^{ -t }{ t }^{ n-1 } ln\left(t\right)} dt .$

To evaluate ${\Gamma}^{'} (1)$, we set $n=1$ thus $\int _{ 0 }^{ \infty }{ { e }^{ -t }ln\left(t\right)} dt .$

We replace ${e}^{-t}$ with $\lim _{ n\rightarrow \infty }{ \left(1-\frac{t}{n} \right)^{n} } .$

Let $s = 1 - \frac{t}{n}$ and $-nds = dt,$ we get $\begin{aligned} \lim _{ n\rightarrow \infty }{ \left[ \int _{ 0 }^{ 1 }{ { s }^{ n } } \left[ ln(n)+ln(1-s) \right] (-n)ds \right] } \\ &= \lim _{ n\rightarrow \infty }{ \left[ nln(n)\int _{ 0 }^{ 1 }{ { s }^{ n } } ds+\int _{ 0 }^{ 1 }{ { s }^{ n } } ln(1-s)ds \right] } \\ &=\lim _{ n\rightarrow \infty }{ \left[ \frac { n }{ n+1 } ln(n)+\frac { -1 }{ n+1 } \int _{ 0 }^{ 1 }{ \frac { { s }^{ n+1 }-1 }{ s-1 } } ds \right] } \\ &=\lim _{ n\rightarrow \infty }{\frac { n }{ n+1 } \left[ ln(n)-{ H }_{ n+1 } \right]} . \end{aligned}$

${H}_{n+1}$ is the harmonic number. By definition $\lim_{n\rightarrow \infty}{\left[{ H }_{ n+1 } -ln(n)\right]}$ is the Euler-Mascheroni constant; therefore, ${\Gamma}^{'} (1) = -\gamma$.