Who derived this equation? [number theory]

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Does anyone know the name of the paper in which this equation first appeared? Thank you!
$$int_0^infty e^-nxx^s-1dx = Pi(s-1)/n^s$$

complex-analysis number-theory reference-request math-history

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edited Jul 31 at 13:44

joriki

164k10179328

asked Jul 31 at 13:36

Christina Daniel

132

add a comment | 

up vote
2
down vote

favorite

1

Does anyone know the name of the paper in which this equation first appeared? Thank you!
$$int_0^infty e^-nxx^s-1dx = Pi(s-1)/n^s$$

complex-analysis number-theory reference-request math-history

share|cite|improve this question

edited Jul 31 at 13:44

joriki

164k10179328

asked Jul 31 at 13:36

Christina Daniel

132

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Does anyone know the name of the paper in which this equation first appeared? Thank you!
$$int_0^infty e^-nxx^s-1dx = Pi(s-1)/n^s$$

complex-analysis number-theory reference-request math-history

share|cite|improve this question

edited Jul 31 at 13:44

joriki

164k10179328

asked Jul 31 at 13:36

Christina Daniel

132

Does anyone know the name of the paper in which this equation first appeared? Thank you!
$$int_0^infty e^-nxx^s-1dx = Pi(s-1)/n^s$$

complex-analysis number-theory reference-request math-history

share|cite|improve this question

edited Jul 31 at 13:44

joriki

164k10179328

asked Jul 31 at 13:36

Christina Daniel

132

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 13:44

joriki

164k10179328

edited Jul 31 at 13:44

joriki

164k10179328

edited Jul 31 at 13:44

joriki

164k10179328

164k10179328

asked Jul 31 at 13:36

Christina Daniel

132

asked Jul 31 at 13:36

Christina Daniel

132

asked Jul 31 at 13:36

Christina Daniel

132

132

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

It was (almost certainly) Euler. For example, we find it in E675, De valoribus integralium a termino variabilis $x=0$ usque ad $x=infty$ extensorum (1781). A translation of this paper may be found on arXiv. At the bottom of p. 3 of the translation one finds

§8. I first put $x=ky$, and because both terms of the integral stay the same, it will be
$$k^n int y^n−1partial y cdot e^−ky= Delta, $$
where this formula is also extended from $y= 0$ up to $y=infty$; then dividing this by $k^n$ we will have
$$ int y^n−1partial y cdot e^−ky=fracDeltak^n $$,where it should however be noted that no negative numbers can be taken for $k$, as otherwise the formula $e^−ky$ would no longer vanish in the case $y=infty$;and these are the only values which ought to be excluded here, so that even imaginary values could be used in place of $k$, and then I pursued these laborious integrations.

(And yes, Euler does consider non-integer $n$ in the paper: he discusses $n=1/2$ later on.)

share|cite|improve this answer

answered Jul 31 at 14:15

Chappers

54.9k74190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much!
  – Christina Daniel
  Jul 31 at 14:31
  

  
  
  
  

add a comment | 

up vote
2
down vote

That formula is easily seen to be equivalent to the integral definition of the Gamma function, introduced by Euler in a letter to Goldbach in 1730. (Note that $Pi(s-1)=Gamma(s)$.)

Source: Wikipedia.

share|cite|improve this answer

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

add a comment | 

up vote
0
down vote

I suppose that it was in Riemann's On the number of primes less than a given magnitude. It's the second formula that appears there.

share|cite|improve this answer

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the question is really about the formula exactly as presented, I suppose you win. ;-)
  – Harald Hanche-Olsen
  Jul 31 at 13:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. So Riemann writes "on making use of the equation" as if the equation is already established, so I am wondering where it came from, or if it is "obvious."
  – Christina Daniel
  Jul 31 at 13:54
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hey! It's Riemann that we're talking about. Of course it was obvious to him.
  – José Carlos Santos
  Jul 31 at 14:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the rest of the paper is more pedagogical... the difference in flow doesn't make sense!
  – Christina Daniel
  Jul 31 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's “obvious” in the sense that it only requires substituting $x=t/n$ in the integral. To Riemann, and most likely to his intended audience, that is no great stretch.
  – Harald Hanche-Olsen
  Jul 31 at 14:13
  
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

It was (almost certainly) Euler. For example, we find it in E675, De valoribus integralium a termino variabilis $x=0$ usque ad $x=infty$ extensorum (1781). A translation of this paper may be found on arXiv. At the bottom of p. 3 of the translation one finds

§8. I first put $x=ky$, and because both terms of the integral stay the same, it will be
$$k^n int y^n−1partial y cdot e^−ky= Delta, $$
where this formula is also extended from $y= 0$ up to $y=infty$; then dividing this by $k^n$ we will have
$$ int y^n−1partial y cdot e^−ky=fracDeltak^n $$,where it should however be noted that no negative numbers can be taken for $k$, as otherwise the formula $e^−ky$ would no longer vanish in the case $y=infty$;and these are the only values which ought to be excluded here, so that even imaginary values could be used in place of $k$, and then I pursued these laborious integrations.

(And yes, Euler does consider non-integer $n$ in the paper: he discusses $n=1/2$ later on.)

share|cite|improve this answer

answered Jul 31 at 14:15

Chappers

54.9k74190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much!
  – Christina Daniel
  Jul 31 at 14:31
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

It was (almost certainly) Euler. For example, we find it in E675, De valoribus integralium a termino variabilis $x=0$ usque ad $x=infty$ extensorum (1781). A translation of this paper may be found on arXiv. At the bottom of p. 3 of the translation one finds

§8. I first put $x=ky$, and because both terms of the integral stay the same, it will be
$$k^n int y^n−1partial y cdot e^−ky= Delta, $$
where this formula is also extended from $y= 0$ up to $y=infty$; then dividing this by $k^n$ we will have
$$ int y^n−1partial y cdot e^−ky=fracDeltak^n $$,where it should however be noted that no negative numbers can be taken for $k$, as otherwise the formula $e^−ky$ would no longer vanish in the case $y=infty$;and these are the only values which ought to be excluded here, so that even imaginary values could be used in place of $k$, and then I pursued these laborious integrations.

(And yes, Euler does consider non-integer $n$ in the paper: he discusses $n=1/2$ later on.)

share|cite|improve this answer

answered Jul 31 at 14:15

Chappers

54.9k74190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much!
  – Christina Daniel
  Jul 31 at 14:31
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

It was (almost certainly) Euler. For example, we find it in E675, De valoribus integralium a termino variabilis $x=0$ usque ad $x=infty$ extensorum (1781). A translation of this paper may be found on arXiv. At the bottom of p. 3 of the translation one finds

§8. I first put $x=ky$, and because both terms of the integral stay the same, it will be
$$k^n int y^n−1partial y cdot e^−ky= Delta, $$
where this formula is also extended from $y= 0$ up to $y=infty$; then dividing this by $k^n$ we will have
$$ int y^n−1partial y cdot e^−ky=fracDeltak^n $$,where it should however be noted that no negative numbers can be taken for $k$, as otherwise the formula $e^−ky$ would no longer vanish in the case $y=infty$;and these are the only values which ought to be excluded here, so that even imaginary values could be used in place of $k$, and then I pursued these laborious integrations.

(And yes, Euler does consider non-integer $n$ in the paper: he discusses $n=1/2$ later on.)

share|cite|improve this answer

answered Jul 31 at 14:15

Chappers

54.9k74190

It was (almost certainly) Euler. For example, we find it in E675, De valoribus integralium a termino variabilis $x=0$ usque ad $x=infty$ extensorum (1781). A translation of this paper may be found on arXiv. At the bottom of p. 3 of the translation one finds

§8. I first put $x=ky$, and because both terms of the integral stay the same, it will be
$$k^n int y^n−1partial y cdot e^−ky= Delta, $$
where this formula is also extended from $y= 0$ up to $y=infty$; then dividing this by $k^n$ we will have
$$ int y^n−1partial y cdot e^−ky=fracDeltak^n $$,where it should however be noted that no negative numbers can be taken for $k$, as otherwise the formula $e^−ky$ would no longer vanish in the case $y=infty$;and these are the only values which ought to be excluded here, so that even imaginary values could be used in place of $k$, and then I pursued these laborious integrations.

(And yes, Euler does consider non-integer $n$ in the paper: he discusses $n=1/2$ later on.)

share|cite|improve this answer

answered Jul 31 at 14:15

Chappers

54.9k74190

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 14:15

Chappers

54.9k74190

answered Jul 31 at 14:15

Chappers

54.9k74190

answered Jul 31 at 14:15

Chappers

54.9k74190

54.9k74190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much!
  – Christina Daniel
  Jul 31 at 14:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much!
  – Christina Daniel
  Jul 31 at 14:31
  

  
  
  
  

Thank you so much!
– Christina Daniel
Jul 31 at 14:31

Thank you so much!
– Christina Daniel
Jul 31 at 14:31

add a comment | 

up vote
2
down vote

That formula is easily seen to be equivalent to the integral definition of the Gamma function, introduced by Euler in a letter to Goldbach in 1730. (Note that $Pi(s-1)=Gamma(s)$.)

Source: Wikipedia.

share|cite|improve this answer

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

add a comment | 

up vote
2
down vote

That formula is easily seen to be equivalent to the integral definition of the Gamma function, introduced by Euler in a letter to Goldbach in 1730. (Note that $Pi(s-1)=Gamma(s)$.)

Source: Wikipedia.

share|cite|improve this answer

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

add a comment | 

up vote
2
down vote

up vote
2
down vote

That formula is easily seen to be equivalent to the integral definition of the Gamma function, introduced by Euler in a letter to Goldbach in 1730. (Note that $Pi(s-1)=Gamma(s)$.)

Source: Wikipedia.

share|cite|improve this answer

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

That formula is easily seen to be equivalent to the integral definition of the Gamma function, introduced by Euler in a letter to Goldbach in 1730. (Note that $Pi(s-1)=Gamma(s)$.)

Source: Wikipedia.

share|cite|improve this answer

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

answered Jul 31 at 13:46

Harald Hanche-Olsen

27.3k23959

27.3k23959

add a comment | 

add a comment | 

up vote
0
down vote

I suppose that it was in Riemann's On the number of primes less than a given magnitude. It's the second formula that appears there.

share|cite|improve this answer

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the question is really about the formula exactly as presented, I suppose you win. ;-)
  – Harald Hanche-Olsen
  Jul 31 at 13:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. So Riemann writes "on making use of the equation" as if the equation is already established, so I am wondering where it came from, or if it is "obvious."
  – Christina Daniel
  Jul 31 at 13:54
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hey! It's Riemann that we're talking about. Of course it was obvious to him.
  – José Carlos Santos
  Jul 31 at 14:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the rest of the paper is more pedagogical... the difference in flow doesn't make sense!
  – Christina Daniel
  Jul 31 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's “obvious” in the sense that it only requires substituting $x=t/n$ in the integral. To Riemann, and most likely to his intended audience, that is no great stretch.
  – Harald Hanche-Olsen
  Jul 31 at 14:13
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

I suppose that it was in Riemann's On the number of primes less than a given magnitude. It's the second formula that appears there.

share|cite|improve this answer

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the question is really about the formula exactly as presented, I suppose you win. ;-)
  – Harald Hanche-Olsen
  Jul 31 at 13:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. So Riemann writes "on making use of the equation" as if the equation is already established, so I am wondering where it came from, or if it is "obvious."
  – Christina Daniel
  Jul 31 at 13:54
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hey! It's Riemann that we're talking about. Of course it was obvious to him.
  – José Carlos Santos
  Jul 31 at 14:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the rest of the paper is more pedagogical... the difference in flow doesn't make sense!
  – Christina Daniel
  Jul 31 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's “obvious” in the sense that it only requires substituting $x=t/n$ in the integral. To Riemann, and most likely to his intended audience, that is no great stretch.
  – Harald Hanche-Olsen
  Jul 31 at 14:13
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

I suppose that it was in Riemann's On the number of primes less than a given magnitude. It's the second formula that appears there.

share|cite|improve this answer

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

I suppose that it was in Riemann's On the number of primes less than a given magnitude. It's the second formula that appears there.

share|cite|improve this answer

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

answered Jul 31 at 13:47

José Carlos Santos

112k1696172

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the question is really about the formula exactly as presented, I suppose you win. ;-)
  – Harald Hanche-Olsen
  Jul 31 at 13:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. So Riemann writes "on making use of the equation" as if the equation is already established, so I am wondering where it came from, or if it is "obvious."
  – Christina Daniel
  Jul 31 at 13:54
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hey! It's Riemann that we're talking about. Of course it was obvious to him.
  – José Carlos Santos
  Jul 31 at 14:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the rest of the paper is more pedagogical... the difference in flow doesn't make sense!
  – Christina Daniel
  Jul 31 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's “obvious” in the sense that it only requires substituting $x=t/n$ in the integral. To Riemann, and most likely to his intended audience, that is no great stretch.
  – Harald Hanche-Olsen
  Jul 31 at 14:13
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the question is really about the formula exactly as presented, I suppose you win. ;-)
  – Harald Hanche-Olsen
  Jul 31 at 13:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. So Riemann writes "on making use of the equation" as if the equation is already established, so I am wondering where it came from, or if it is "obvious."
  – Christina Daniel
  Jul 31 at 13:54
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hey! It's Riemann that we're talking about. Of course it was obvious to him.
  – José Carlos Santos
  Jul 31 at 14:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But the rest of the paper is more pedagogical... the difference in flow doesn't make sense!
  – Christina Daniel
  Jul 31 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's “obvious” in the sense that it only requires substituting $x=t/n$ in the integral. To Riemann, and most likely to his intended audience, that is no great stretch.
  – Harald Hanche-Olsen
  Jul 31 at 14:13
  
  

  
  
  
  

If the question is really about the formula exactly as presented, I suppose you win. ;-)
– Harald Hanche-Olsen
Jul 31 at 13:49

If the question is really about the formula exactly as presented, I suppose you win. ;-)
– Harald Hanche-Olsen
Jul 31 at 13:49

Yes. So Riemann writes "on making use of the equation" as if the equation is already established, so I am wondering where it came from, or if it is "obvious."
– Christina Daniel
Jul 31 at 13:54

Yes. So Riemann writes "on making use of the equation" as if the equation is already established, so I am wondering where it came from, or if it is "obvious."
– Christina Daniel
Jul 31 at 13:54

1

1

Hey! It's Riemann that we're talking about. Of course it was obvious to him.
– José Carlos Santos
Jul 31 at 14:00

Hey! It's Riemann that we're talking about. Of course it was obvious to him.
– José Carlos Santos
Jul 31 at 14:00

But the rest of the paper is more pedagogical... the difference in flow doesn't make sense!
– Christina Daniel
Jul 31 at 14:12

But the rest of the paper is more pedagogical... the difference in flow doesn't make sense!
– Christina Daniel
Jul 31 at 14:12

1

1

It's “obvious” in the sense that it only requires substituting $x=t/n$ in the integral. To Riemann, and most likely to his intended audience, that is no great stretch.
– Harald Hanche-Olsen
Jul 31 at 14:13

It's “obvious” in the sense that it only requires substituting $x=t/n$ in the integral. To Riemann, and most likely to his intended audience, that is no great stretch.
– Harald Hanche-Olsen
Jul 31 at 14:13

add a comment | 

 

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