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 \[ \begin{align*} \lim _{ n\rightarrow \infty }{ \left[ \int _{ 0 }^{ 1 }{ { s }^{ n } } \left[ ln(n)+ln(1-s) \right] (-n)ds \right] } \\ &= \lim _{ n\rightarrow \infty }{ \left[ nln(n)\int _{ 0 }^{ 1 }{ { s }^{ n } } ds+\int _{ 0 }^{ 1 }{ { s }^{ n } } ln(1-s)ds \right] } \\ &=\lim _{ n\rightarrow \infty }{ \left[ \frac { n }{ n+1 } ln(n)+\frac { -1 }{ n+1 } \int _{ 0 }^{ 1 }{ \frac { { s }^{ n+1 }-1 }{ s-1 } } ds \right] } \\ &=\lim _{ n\rightarrow \infty }{\frac { n }{ n+1 } \left[ ln(n)-{ H }_{ n+1 } \right]} . \end{align*}\]

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

Note by Steven Zheng
3 years, 5 months ago

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I just noticed I wrote a note with the exact title. Might as well move the content to that note.

Steven Zheng - 3 years, 5 months ago

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How do u add alignment?

Aneesh Kundu - 3 years, 5 months ago

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Consider this page: LaTex Align equations

Steven Zheng - 3 years, 5 months ago

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Thnx

Aneesh Kundu - 3 years, 5 months ago

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