Euler-Mascheroni Constant

## **Problem**

**Part 1:** Differentiate the gamma function

Γ
(
n
# )

∫

0

∞

t

n
−
1

e

−
t

d
t

{\displaystyle \Gamma(n) = \int \_{ 0 }^{ \infty }{ { t }^{ n-1 }{ e }^{ -t } } dt }

.
**Part 2:** The Euler-Mascheroni constant is defined as

lim

n
→
∞

[

H

n
1

−
ln
⁡

(
n
)

]

=
γ

{\displaystyle \lim\_{n\rightarrow \infty}{\left[{ H }\_{ n+1 } -\ln{(n)}\right]} = \gamma }

.
Use the result in part 1 to show that

Γ

′

(
1
# )
−
γ

{\displaystyle {\Gamma}^{'} (1) = -\gamma }

. For reference, the numerical value of the Euler-Mascheroni constant is roughly 0.577216.

## **Solution**

**Part 1**

We begin with the integral definition of the Gamma function

lim

x
→
∞

∫

0

x

e

−
t

t

n
−
1

d
t

{\displaystyle \lim \_{ x\rightarrow \infty }{ \int \_{ 0 }^{ x }{ { e }^{ -t }{ t }^{ n-1 } } dt } }

.
To differentiate under the integral, use Leibniz Rule:

d

d
x

∫

a
(
x
)

b
(
x
)

f
(
x
,
t
)
d
# t

∫

a
(
x
)

b
(
x
)

∂

∂
x

f
(
x
,
t
)
d
t
f

(

x
,
b
(
x
)

)

⋅

b
′

(
x
)
−
f

(

x
,
a
(
x
)

)

⋅

a
′

(
x
)

{\displaystyle \frac{d}{dx}\int\_{a(x)}^{b(x)} f(x,t)dt = \int\_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) dt + f\big(x,b(x)\big)\cdot b'(x) - f\big(x,a(x)\big)\cdot a'(x) }

.
For this problem, replace

x

{\displaystyle x}

 with

n

{\displaystyle n}

:

∂
f

∂
n

=

t

n
−
1

e

−
t

ln
⁡

(
t
)

{\displaystyle \frac{\partial f}{\partial n} = {t}^{n-1}{e}^{-t} \ln{(t)} }

.
Thus, we evaluate the limit

Γ
′

(
n
,
t
)

=

lim

x
→
∞

[

∫

0

x

t

n
−
1

ln
⁡
(
t
)

e

−
t

d
t

(

x

n
−
1

e

−
x

)

⋅
1
−
0
⋅
0

]

=

lim

x
→
∞

[

∫

0

x

t

n
−
1

ln
⁡
(
t
)

e

−
t

d
t

x

n
−
1

e

−
x

]

=

∫

0

∞

e

−
t

t

n
−
1

ln
⁡

(
t
)

d
t

{\displaystyle \begin{align}
\Gamma '(n,t) &= \lim \_{ x\rightarrow \infty }{ \left[\int \_{ 0 }^{ x }{{t}^{n-1}\ln(t){e}^{-t}} dt + \left({ x }^{ n-1 }{ e }^{-x}\right) \cdot 1 - 0 \cdot 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{align}}

**Part 2**

To evaluate

Γ

′

(
1
)

{\displaystyle {\Gamma}^{'} (1) }

, set

# n
1

{\displaystyle n=1}

, thus

∫

0

∞

e

−
t

ln
⁡

(
t
)

d
t

{\displaystyle \int \_{ 0 }^{ \infty }{ { e }^{ -t }\ln\left(t\right)} dt }

.
We replace

e

−
t

{\displaystyle {e}^{-t} }

 with

lim

n
→
∞

(

1
−

t
n

)

n

{\displaystyle \lim \_{ n\rightarrow \infty }{ \left(1-\frac{t}{n} \right)^{n} } }

.

Let

# s
1
−

t
n

{\displaystyle s = 1 - \frac{t}{n} }

 and

−
n
d
# s
d
t

{\displaystyle -nds = dt }

, we get

lim

n
→
∞

[

∫

1

0

s

n

[

ln
⁡
(
n
)
ln
⁡
(
1
−
s
)

]

(
−
n
)
d
s

]

=

lim

n
→
∞

[

n
ln
⁡

(
n
)

∫

0

1

s

n

d
s

∫

0

1

s

n

ln
⁡
(
1
−
s
)
d
s

]

=

lim

n
→
∞

[

n

n
1

ln
⁡

(
n
)

−
1

n
1

∫

0

1

s

n
1

−
1

s
−
1

d
s

]

=

lim

n
→
∞

n

n
1

[

ln
⁡

(
n
)

−

H

n
1

]

{\displaystyle \begin{align}
\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[ n\ln{(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} }

Since the Euler-Mascheroni constant is defined as

lim

n
→
∞

[

H

n
1

−
ln
⁡

(
n
)

]

{\displaystyle \lim\_{n\rightarrow \infty}{\left[{ H }\_{ n+1 } -\ln{(n)} \right]} }

we discover that

Γ

′

(
1
# )
−
γ

{\displaystyle {\Gamma}^{'} (1) = -\gamma }

.