Derivative of the Gamma Function

Problem 1. Differentiate the gamma function Γ(n)=0tn1etdt.\Gamma(n) = \int _{ 0 }^{ \infty }{ { t }^{ n-1 }{ e }^{ -t } } dt.

Solution

We begin with the integral definition of the Gamma function limx0xettn1dt.\lim _{ x\rightarrow \infty }{ \int _{ 0 }^{ x }{ { e }^{ -t }{ t }^{ n-1 } } dt } .

To differentiate under the integral, we use Leibniz Rule .

First we find fn=tn1etln(t).\frac{\partial f}{\partial n} = {t}^{n-1}{e}^{-t} ln\left(t\right).

Then we evaluate the limit

Γ(n,t)=limx[0xtn1(lnt)etdtxn1ex0]=limx[0xtn1(lnt)etdtxn1ex]=0ettn1ln(t)dt. \begin{aligned} \Gamma '(n,t) &= \lim _{ x\rightarrow \infty }{ \left[\int _{ 0 }^{ x }{{t}^{n-1}(ln t){e}^{-t}} dt - { x }^{ n-1 }{ e }^{-x} - 0 \right] } \\ &= \lim _{ x\rightarrow \infty }{ \left[\int _{ 0 }^{ x }{{t}^{n-1}(ln t){e}^{-t}} dt - { x }^{ n-1 }{ e }^{-x}\right] } \\ &= \int _{ 0 }^{ \infty }{ { e }^{ -t }{ t }^{ n-1 } ln\left(t\right)} dt . \end{aligned}

Problem 2. Show that Γ(1)=γ{\Gamma}^{'} (1) = -\gamma where γ\gamma is the Euler-Mascheroni constant.

Solution

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

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

Let s=1tns = 1 - \frac{t}{n} and nds=dt,-nds = dt, we get limn[10sn[ln(n)+ln(1s)](n)ds]=limn[nln(n)01snds+01snln(1s)ds]=limn[nn+1ln(n)+1n+101sn+11s1ds]=limnnn+1[ln(n)Hn+1]. \begin{aligned} \lim _{ n\rightarrow \infty }{ \left[ \int _{ 1}^{ 0}{ { 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}

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

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
5 years ago

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Why would you write Γ(n,t)\Gamma(n,t) ? the tt is a dummy variable.

Haroun Meghaichi - 5 years ago

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We don't?

Steven Zheng - 5 years ago

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@Steven Zheng Yes, you probably got mixed up with the incomplete gamma function.

Haroun Meghaichi - 5 years ago

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In the end, the function is only a function of n, and not t. Like Mr./Ms. Haroun said, it's a dummy variable.

Aditya Ranjan - 2 months, 1 week ago

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Hello, How did you integrate 01snln(1s)ds?\int_0^1 s^n ln(1-s)ds?

Winod DHAMNEKAR - 6 months, 2 weeks ago

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