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Prove that $int_0^infty e^-x ln x d x = - gamma $

Evaluate $int_0^infty fraclog(1+x^3)(1+x^2)^2dx$ and $int_0^infty fraclog(1+x^4)(1+x^2)^2dx$Evaluating $int_0^infty fracdx1+x^4$.Prove that $int_0^infty y^frac12 e^-y^2 int_0^infty y^-frac12 e^-y^2 =fracpi2^frac32$Convergence of $int_0 ^inftyfraccos tt^alpha dt$ related to $Gamma$ functionEvaluating integral $displaystyle int_0^infty x^n e^frac-xn dx$ with Gamma-function?Show that $int_0^1(lnGamma)(x)mathrm dx=ln(sqrt2pi)$Prove that $int_0^infty e^-x^4 dx = Gammaleft(frac54right) = left(frac14right)!$ using the gamma functionHow to integrate gamma functionIs there another way of evaluating $lim_x to 0 Gamma(x)(gamma+psi(1+x))=fracpi^26$An $operatornameerfi(x)e^-x^2$ integral

4

$begingroup$

I can see it is right by using some knowledge of the Gamma function. We have

$$ Gamma(alpha ) = int_0^infty e^-x x^alpha - 1 dx . $$

Differentiating with respect to $alpha$, we get

$$ fracdGammad alpha = int_0^infty e^-x x^alpha - 1 ln x dx . $$

Setting $alpha = 1 $, we get

$$ int_0^infty e^-x ln x dx = fracdGammadalpha bigg|_alpha = 1 = - gamma. $$

But the knowledge $d Gamma/d alpha |_alpha = 1=-gamma$ is a mystery to me.

Can anyone find an elementary proof?

integration improper-integrals gamma-function

share|cite|improve this question

asked Mar 15 at 5:15

JohnJohn

1874

$endgroup$

add a comment | 

4

$begingroup$

I can see it is right by using some knowledge of the Gamma function. We have

$$ Gamma(alpha ) = int_0^infty e^-x x^alpha - 1 dx . $$

Differentiating with respect to $alpha$, we get

$$ fracdGammad alpha = int_0^infty e^-x x^alpha - 1 ln x dx . $$

Setting $alpha = 1 $, we get

$$ int_0^infty e^-x ln x dx = fracdGammadalpha bigg|_alpha = 1 = - gamma. $$

But the knowledge $d Gamma/d alpha |_alpha = 1=-gamma$ is a mystery to me.

Can anyone find an elementary proof?

integration improper-integrals gamma-function

share|cite|improve this question

asked Mar 15 at 5:15

JohnJohn

1874

$endgroup$

add a comment | 

$begingroup$

I can see it is right by using some knowledge of the Gamma function. We have

$$ Gamma(alpha ) = int_0^infty e^-x x^alpha - 1 dx . $$

Differentiating with respect to $alpha$, we get

$$ fracdGammad alpha = int_0^infty e^-x x^alpha - 1 ln x dx . $$

Setting $alpha = 1 $, we get

$$ int_0^infty e^-x ln x dx = fracdGammadalpha bigg|_alpha = 1 = - gamma. $$

But the knowledge $d Gamma/d alpha |_alpha = 1=-gamma$ is a mystery to me.

Can anyone find an elementary proof?

integration improper-integrals gamma-function

share|cite|improve this question

asked Mar 15 at 5:15

JohnJohn

1874

$endgroup$

share|cite|improve this question

asked Mar 15 at 5:15

JohnJohn

1874

share|cite|improve this question

asked Mar 15 at 5:15

JohnJohn

1874

asked Mar 15 at 5:15

JohnJohn

1874

asked Mar 15 at 5:15

JohnJohn

1874

3 Answers
3

active

oldest

votes

5

$begingroup$

You can prove that
$$lim I_n=int_0^infty e^-xln x,dx$$
using the dominated convergence theorem, where
$$I_n=int_0^nleft(1-frac xnright)^nln x,dx.$$
Now substitute $y=1-x/n$. Then
beginalign
I_n&=nint_0^1y^nln(n(1-y)),dy\
&=nln nint_0^1 y^n,dy+nint_0^1y^nln(1-y),dy\
&=frac nn+1left(ln n-int_0^1frac1-y^n+11-y,dyright)\
&=frac nn+1left(ln n-sum_k=1^n+1frac1kright)
endalign
where we have integrated by parts along the way. This tends to $-gamma$.

share|cite|improve this answer

answered Mar 15 at 5:34

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you write $y^n dy = d (y^n+1 -1 ) /(n+1)$ in the second integral in the second line?
  $endgroup$
  – John
  Mar 15 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @John In effect, to avoid a singularity at $1$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 15 at 19:07
  

  
  
  
  

add a comment | 

2

$begingroup$

Here we will address your integral:
beginequation
I = int_0^infty e^-x ln(x):dx nonumber
endequation

To do so we use the fact that:
beginequation
lim_a rightarrow 0^+ fracpartialpartial a x^a = ln(x)
endequation

As such (and by Leibniz's Integral Rule):
beginalign
I &= int_0^infty e^-x ln(x):dx = lim_a rightarrow 0^+ fracpartialpartial aint_0^infty e^-x x^a:dx = lim_a rightarrow 0^+ fracpartialpartial a Gamma(a + 1) nonumber \
&= lim_a rightarrow 0^+ Gamma'(a + 1) = lim_a rightarrow 0^+ Gamma(a + 1):psi^(0)(a + 1)nonumber \
&= Gamma(1):psi^(0)(0) = psi^(0)(0) = -gamma
endalign

share|cite|improve this answer

answered Mar 15 at 5:35

DavidGDavidG

1

$endgroup$

add a comment | 

0

$begingroup$

Your method seems like a reasonable way to derive this result, and you are almost there. Now use the Taylor expansion of the digamma function, which is:

$$psi(z+1) = -gamma - sum_k=1^infty zeta(k+1) cdot (-z)^k.$$

Substituting $z=0$ gives:

$$psi(1) = -gamma - sum_k=1^infty zeta(k+1) cdot 0^k = -gamma.$$

So you have:

$$int_0^infty e^-x ln x dx = Gamma'(1) = psi(1) cdot Gamma(1) = -gamma cdot 1 = -gamma.$$

share|cite|improve this answer

answered Mar 15 at 5:33

BenBen

1,840215

$endgroup$

add a comment | 

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5

$begingroup$

You can prove that
$$lim I_n=int_0^infty e^-xln x,dx$$
using the dominated convergence theorem, where
$$I_n=int_0^nleft(1-frac xnright)^nln x,dx.$$
Now substitute $y=1-x/n$. Then
beginalign
I_n&=nint_0^1y^nln(n(1-y)),dy\
&=nln nint_0^1 y^n,dy+nint_0^1y^nln(1-y),dy\
&=frac nn+1left(ln n-int_0^1frac1-y^n+11-y,dyright)\
&=frac nn+1left(ln n-sum_k=1^n+1frac1kright)
endalign
where we have integrated by parts along the way. This tends to $-gamma$.

share|cite|improve this answer

answered Mar 15 at 5:34

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you write $y^n dy = d (y^n+1 -1 ) /(n+1)$ in the second integral in the second line?
  $endgroup$
  – John
  Mar 15 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @John In effect, to avoid a singularity at $1$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 15 at 19:07
  

  
  
  
  

add a comment | 

5

$begingroup$

You can prove that
$$lim I_n=int_0^infty e^-xln x,dx$$
using the dominated convergence theorem, where
$$I_n=int_0^nleft(1-frac xnright)^nln x,dx.$$
Now substitute $y=1-x/n$. Then
beginalign
I_n&=nint_0^1y^nln(n(1-y)),dy\
&=nln nint_0^1 y^n,dy+nint_0^1y^nln(1-y),dy\
&=frac nn+1left(ln n-int_0^1frac1-y^n+11-y,dyright)\
&=frac nn+1left(ln n-sum_k=1^n+1frac1kright)
endalign
where we have integrated by parts along the way. This tends to $-gamma$.

share|cite|improve this answer

answered Mar 15 at 5:34

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you write $y^n dy = d (y^n+1 -1 ) /(n+1)$ in the second integral in the second line?
  $endgroup$
  – John
  Mar 15 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @John In effect, to avoid a singularity at $1$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 15 at 19:07
  

  
  
  
  

add a comment | 

$begingroup$

You can prove that
$$lim I_n=int_0^infty e^-xln x,dx$$
using the dominated convergence theorem, where
$$I_n=int_0^nleft(1-frac xnright)^nln x,dx.$$
Now substitute $y=1-x/n$. Then
beginalign
I_n&=nint_0^1y^nln(n(1-y)),dy\
&=nln nint_0^1 y^n,dy+nint_0^1y^nln(1-y),dy\
&=frac nn+1left(ln n-int_0^1frac1-y^n+11-y,dyright)\
&=frac nn+1left(ln n-sum_k=1^n+1frac1kright)
endalign
where we have integrated by parts along the way. This tends to $-gamma$.

share|cite|improve this answer

answered Mar 15 at 5:34

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

share|cite|improve this answer

answered Mar 15 at 5:34

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered Mar 15 at 5:34

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered Mar 15 at 5:34

Lord Shark the UnknownLord Shark the Unknown

106k1161133

2

$begingroup$

Here we will address your integral:
beginequation
I = int_0^infty e^-x ln(x):dx nonumber
endequation

To do so we use the fact that:
beginequation
lim_a rightarrow 0^+ fracpartialpartial a x^a = ln(x)
endequation

As such (and by Leibniz's Integral Rule):
beginalign
I &= int_0^infty e^-x ln(x):dx = lim_a rightarrow 0^+ fracpartialpartial aint_0^infty e^-x x^a:dx = lim_a rightarrow 0^+ fracpartialpartial a Gamma(a + 1) nonumber \
&= lim_a rightarrow 0^+ Gamma'(a + 1) = lim_a rightarrow 0^+ Gamma(a + 1):psi^(0)(a + 1)nonumber \
&= Gamma(1):psi^(0)(0) = psi^(0)(0) = -gamma
endalign

share|cite|improve this answer

answered Mar 15 at 5:35

DavidGDavidG

1

$endgroup$

add a comment | 

2

$begingroup$

Here we will address your integral:
beginequation
I = int_0^infty e^-x ln(x):dx nonumber
endequation

To do so we use the fact that:
beginequation
lim_a rightarrow 0^+ fracpartialpartial a x^a = ln(x)
endequation

As such (and by Leibniz's Integral Rule):
beginalign
I &= int_0^infty e^-x ln(x):dx = lim_a rightarrow 0^+ fracpartialpartial aint_0^infty e^-x x^a:dx = lim_a rightarrow 0^+ fracpartialpartial a Gamma(a + 1) nonumber \
&= lim_a rightarrow 0^+ Gamma'(a + 1) = lim_a rightarrow 0^+ Gamma(a + 1):psi^(0)(a + 1)nonumber \
&= Gamma(1):psi^(0)(0) = psi^(0)(0) = -gamma
endalign

share|cite|improve this answer

answered Mar 15 at 5:35

DavidGDavidG

1

$endgroup$

add a comment | 

$begingroup$

Here we will address your integral:
beginequation
I = int_0^infty e^-x ln(x):dx nonumber
endequation

To do so we use the fact that:
beginequation
lim_a rightarrow 0^+ fracpartialpartial a x^a = ln(x)
endequation

As such (and by Leibniz's Integral Rule):
beginalign
I &= int_0^infty e^-x ln(x):dx = lim_a rightarrow 0^+ fracpartialpartial aint_0^infty e^-x x^a:dx = lim_a rightarrow 0^+ fracpartialpartial a Gamma(a + 1) nonumber \
&= lim_a rightarrow 0^+ Gamma'(a + 1) = lim_a rightarrow 0^+ Gamma(a + 1):psi^(0)(a + 1)nonumber \
&= Gamma(1):psi^(0)(0) = psi^(0)(0) = -gamma
endalign

share|cite|improve this answer

answered Mar 15 at 5:35

DavidGDavidG

1

$endgroup$

share|cite|improve this answer

answered Mar 15 at 5:35

DavidGDavidG

1

answered Mar 15 at 5:35

DavidGDavidG

1

answered Mar 15 at 5:35

DavidGDavidG

1

0

$begingroup$

Your method seems like a reasonable way to derive this result, and you are almost there. Now use the Taylor expansion of the digamma function, which is:

$$psi(z+1) = -gamma - sum_k=1^infty zeta(k+1) cdot (-z)^k.$$

Substituting $z=0$ gives:

$$psi(1) = -gamma - sum_k=1^infty zeta(k+1) cdot 0^k = -gamma.$$

So you have:

$$int_0^infty e^-x ln x dx = Gamma'(1) = psi(1) cdot Gamma(1) = -gamma cdot 1 = -gamma.$$

share|cite|improve this answer

answered Mar 15 at 5:33

BenBen

1,840215

$endgroup$

add a comment | 

0

$begingroup$

Your method seems like a reasonable way to derive this result, and you are almost there. Now use the Taylor expansion of the digamma function, which is:

$$psi(z+1) = -gamma - sum_k=1^infty zeta(k+1) cdot (-z)^k.$$

Substituting $z=0$ gives:

$$psi(1) = -gamma - sum_k=1^infty zeta(k+1) cdot 0^k = -gamma.$$

So you have:

$$int_0^infty e^-x ln x dx = Gamma'(1) = psi(1) cdot Gamma(1) = -gamma cdot 1 = -gamma.$$

share|cite|improve this answer

answered Mar 15 at 5:33

BenBen

1,840215

$endgroup$

add a comment | 

$begingroup$

Your method seems like a reasonable way to derive this result, and you are almost there. Now use the Taylor expansion of the digamma function, which is:

$$psi(z+1) = -gamma - sum_k=1^infty zeta(k+1) cdot (-z)^k.$$

Substituting $z=0$ gives:

$$psi(1) = -gamma - sum_k=1^infty zeta(k+1) cdot 0^k = -gamma.$$

So you have:

$$int_0^infty e^-x ln x dx = Gamma'(1) = psi(1) cdot Gamma(1) = -gamma cdot 1 = -gamma.$$

share|cite|improve this answer

answered Mar 15 at 5:33

BenBen

1,840215

$endgroup$

share|cite|improve this answer

answered Mar 15 at 5:33

BenBen

1,840215

answered Mar 15 at 5:33

BenBen

1,840215

answered Mar 15 at 5:33

BenBen

1,840215