# 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, ${\Gamma}^{'} (1) = -\gamma$. Note by Steven Zheng
5 years, 10 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

- 5 years, 10 months ago

- 5 years, 10 months ago

- 5 years, 10 months ago

Thnx

- 5 years, 10 months ago