Evaluating a nested log integral

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Question:$$intlimits_0^1mathrm dx,frac loglogfrac 1x(1+x)^2=frac 12logfrac pi2-frac gamma2$$

I’ve had some practice with similar integrals, but this one eludes me for some reason. I first made the transformation $xmapsto-log x$ to get rid of the nested log. Therefore$$mathfrakI=intlimits_0^inftymathrm dx,frac e^-xlog x(1+e^-x)^2$$
The inside integrand can be rewritten as an infinite series to get$$mathfrakI=sumlimits_ngeq0(n+1)(-1)^nintlimits_0^inftymathrm dx, e^-x(n+1)log x$$The inside integral, I thought, could be evaluated by differentiating the gamma function to get$$intlimits_0^inftymathrm dt, e^-t(n+1)log t=-frac gamman+1-frac log(n+1)n+1$$
However, when I simplify everything and split the sum, neither sum converges. If we consider it as a Cesaro sum, then I know for sure that$$sumlimits_ngeq0(-1)^n=frac 12$$Which eventually does give the right answer. But I’m not sure if we’re quite allowed to do that especially because in a general sense, neither sum converges.

integration

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asked Jul 16 at 20:18

Frank

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Question:$$intlimits_0^1mathrm dx,frac loglogfrac 1x(1+x)^2=frac 12logfrac pi2-frac gamma2$$

I’ve had some practice with similar integrals, but this one eludes me for some reason. I first made the transformation $xmapsto-log x$ to get rid of the nested log. Therefore$$mathfrakI=intlimits_0^inftymathrm dx,frac e^-xlog x(1+e^-x)^2$$
The inside integrand can be rewritten as an infinite series to get$$mathfrakI=sumlimits_ngeq0(n+1)(-1)^nintlimits_0^inftymathrm dx, e^-x(n+1)log x$$The inside integral, I thought, could be evaluated by differentiating the gamma function to get$$intlimits_0^inftymathrm dt, e^-t(n+1)log t=-frac gamman+1-frac log(n+1)n+1$$
However, when I simplify everything and split the sum, neither sum converges. If we consider it as a Cesaro sum, then I know for sure that$$sumlimits_ngeq0(-1)^n=frac 12$$Which eventually does give the right answer. But I’m not sure if we’re quite allowed to do that especially because in a general sense, neither sum converges.

integration

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asked Jul 16 at 20:18

Frank

3,7051629

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Question:$$intlimits_0^1mathrm dx,frac loglogfrac 1x(1+x)^2=frac 12logfrac pi2-frac gamma2$$

I’ve had some practice with similar integrals, but this one eludes me for some reason. I first made the transformation $xmapsto-log x$ to get rid of the nested log. Therefore$$mathfrakI=intlimits_0^inftymathrm dx,frac e^-xlog x(1+e^-x)^2$$
The inside integrand can be rewritten as an infinite series to get$$mathfrakI=sumlimits_ngeq0(n+1)(-1)^nintlimits_0^inftymathrm dx, e^-x(n+1)log x$$The inside integral, I thought, could be evaluated by differentiating the gamma function to get$$intlimits_0^inftymathrm dt, e^-t(n+1)log t=-frac gamman+1-frac log(n+1)n+1$$
However, when I simplify everything and split the sum, neither sum converges. If we consider it as a Cesaro sum, then I know for sure that$$sumlimits_ngeq0(-1)^n=frac 12$$Which eventually does give the right answer. But I’m not sure if we’re quite allowed to do that especially because in a general sense, neither sum converges.

integration

share|cite|improve this question

asked Jul 16 at 20:18

Frank

3,7051629

Question:$$intlimits_0^1mathrm dx,frac loglogfrac 1x(1+x)^2=frac 12logfrac pi2-frac gamma2$$

I’ve had some practice with similar integrals, but this one eludes me for some reason. I first made the transformation $xmapsto-log x$ to get rid of the nested log. Therefore$$mathfrakI=intlimits_0^inftymathrm dx,frac e^-xlog x(1+e^-x)^2$$
The inside integrand can be rewritten as an infinite series to get$$mathfrakI=sumlimits_ngeq0(n+1)(-1)^nintlimits_0^inftymathrm dx, e^-x(n+1)log x$$The inside integral, I thought, could be evaluated by differentiating the gamma function to get$$intlimits_0^inftymathrm dt, e^-t(n+1)log t=-frac gamman+1-frac log(n+1)n+1$$
However, when I simplify everything and split the sum, neither sum converges. If we consider it as a Cesaro sum, then I know for sure that$$sumlimits_ngeq0(-1)^n=frac 12$$Which eventually does give the right answer. But I’m not sure if we’re quite allowed to do that especially because in a general sense, neither sum converges.

integration

share|cite|improve this question

asked Jul 16 at 20:18

Frank

3,7051629

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 20:18

Frank

3,7051629

asked Jul 16 at 20:18

Frank

3,7051629

asked Jul 16 at 20:18

Frank

3,7051629

3,7051629

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By the dominated convergence theorem we have
$$ mathfrakI = lim_r nearrow 1 I(r) , ,$$
where for $r in (0,1)$ we have defined
$$ I(r) = intlimits_0^1 mathrmd x,frac loglogfrac 1x(1+r x)^2 , . $$
With this regularisation interchanging summation and integration is actually justified and your calculations lead to
$$ I(r) = - gamma sum limits_n=0^infty (-r)^n - sum limits_n=0^infty (-r)^n log(1+n) equiv I_1 (r) + I_2(r) , . $$
The first sum is easy:
$$ I_1(r) = - fracgamma1+r , , $$
so $lim_r nearrow 1 I_1(r) = - fracgamma2$ .

For the second sum we can write
beginalign
I_2(r) &= frac1r sum_n=1^infty (-r)^n log(n) \
&= frac12r sum_k=1^infty [2 r^2k log(2k) - r^2k-1 log(2k-1) - r^2k+1 log(2k+1)] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r) log(2k+1) - frac1r (1-r) log(2k-1)right] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r)^2 log(2k+1)right] , .
endalign
The second term can be estimated by
beginalign
frac(1-r)^22r sum_k=1^infty r^2k log(2k+1) &leq frac(1-r)^22r^2 sum_n=1^infty sqrtn r^n \
&= frac(1-r)^22r^2 operatornameLi_-1/2 (r) \
&= fracsqrtpi4 r^2 sqrt1-r + mathcalO left((1-r)^3/2right)
endalign
as $r nearrow 1$. The asymptotic behaviour of the polylogarithm can be deduced from the series given here (the second one below 2.). Now we can use the monotone convergence theorem and Wallis' product to find
$$ lim_r nearrow 1 I_2 (r) = frac12 sum_k=1^infty logleft(frac4k^24k^2-1right) = frac12 log left(prod_k=1^infty frac4k^24k^2-1right) = frac12 log left(fracpi2right) , . $$

Therefore
$$ mathfrakI = frac12 left[log left(fracpi2right) - gammaright]$$
as claimed.

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answered Jul 16 at 21:32

ComplexYetTrivial

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  Confirmed by Mathematica: $frac12 left(log left(fracpi 2right)-gamma right)$
  – David G. Stork
  Jul 16 at 21:51
  

  
  
  
  

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1 Answer
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1 Answer
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By the dominated convergence theorem we have
$$ mathfrakI = lim_r nearrow 1 I(r) , ,$$
where for $r in (0,1)$ we have defined
$$ I(r) = intlimits_0^1 mathrmd x,frac loglogfrac 1x(1+r x)^2 , . $$
With this regularisation interchanging summation and integration is actually justified and your calculations lead to
$$ I(r) = - gamma sum limits_n=0^infty (-r)^n - sum limits_n=0^infty (-r)^n log(1+n) equiv I_1 (r) + I_2(r) , . $$
The first sum is easy:
$$ I_1(r) = - fracgamma1+r , , $$
so $lim_r nearrow 1 I_1(r) = - fracgamma2$ .

For the second sum we can write
beginalign
I_2(r) &= frac1r sum_n=1^infty (-r)^n log(n) \
&= frac12r sum_k=1^infty [2 r^2k log(2k) - r^2k-1 log(2k-1) - r^2k+1 log(2k+1)] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r) log(2k+1) - frac1r (1-r) log(2k-1)right] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r)^2 log(2k+1)right] , .
endalign
The second term can be estimated by
beginalign
frac(1-r)^22r sum_k=1^infty r^2k log(2k+1) &leq frac(1-r)^22r^2 sum_n=1^infty sqrtn r^n \
&= frac(1-r)^22r^2 operatornameLi_-1/2 (r) \
&= fracsqrtpi4 r^2 sqrt1-r + mathcalO left((1-r)^3/2right)
endalign
as $r nearrow 1$. The asymptotic behaviour of the polylogarithm can be deduced from the series given here (the second one below 2.). Now we can use the monotone convergence theorem and Wallis' product to find
$$ lim_r nearrow 1 I_2 (r) = frac12 sum_k=1^infty logleft(frac4k^24k^2-1right) = frac12 log left(prod_k=1^infty frac4k^24k^2-1right) = frac12 log left(fracpi2right) , . $$

Therefore
$$ mathfrakI = frac12 left[log left(fracpi2right) - gammaright]$$
as claimed.

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answered Jul 16 at 21:32

ComplexYetTrivial

2,702624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by Mathematica: $frac12 left(log left(fracpi 2right)-gamma right)$
  – David G. Stork
  Jul 16 at 21:51
  

  
  
  
  

add a comment | 

up vote
5
down vote

By the dominated convergence theorem we have
$$ mathfrakI = lim_r nearrow 1 I(r) , ,$$
where for $r in (0,1)$ we have defined
$$ I(r) = intlimits_0^1 mathrmd x,frac loglogfrac 1x(1+r x)^2 , . $$
With this regularisation interchanging summation and integration is actually justified and your calculations lead to
$$ I(r) = - gamma sum limits_n=0^infty (-r)^n - sum limits_n=0^infty (-r)^n log(1+n) equiv I_1 (r) + I_2(r) , . $$
The first sum is easy:
$$ I_1(r) = - fracgamma1+r , , $$
so $lim_r nearrow 1 I_1(r) = - fracgamma2$ .

For the second sum we can write
beginalign
I_2(r) &= frac1r sum_n=1^infty (-r)^n log(n) \
&= frac12r sum_k=1^infty [2 r^2k log(2k) - r^2k-1 log(2k-1) - r^2k+1 log(2k+1)] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r) log(2k+1) - frac1r (1-r) log(2k-1)right] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r)^2 log(2k+1)right] , .
endalign
The second term can be estimated by
beginalign
frac(1-r)^22r sum_k=1^infty r^2k log(2k+1) &leq frac(1-r)^22r^2 sum_n=1^infty sqrtn r^n \
&= frac(1-r)^22r^2 operatornameLi_-1/2 (r) \
&= fracsqrtpi4 r^2 sqrt1-r + mathcalO left((1-r)^3/2right)
endalign
as $r nearrow 1$. The asymptotic behaviour of the polylogarithm can be deduced from the series given here (the second one below 2.). Now we can use the monotone convergence theorem and Wallis' product to find
$$ lim_r nearrow 1 I_2 (r) = frac12 sum_k=1^infty logleft(frac4k^24k^2-1right) = frac12 log left(prod_k=1^infty frac4k^24k^2-1right) = frac12 log left(fracpi2right) , . $$

Therefore
$$ mathfrakI = frac12 left[log left(fracpi2right) - gammaright]$$
as claimed.

share|cite|improve this answer

answered Jul 16 at 21:32

ComplexYetTrivial

2,702624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by Mathematica: $frac12 left(log left(fracpi 2right)-gamma right)$
  – David G. Stork
  Jul 16 at 21:51
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

By the dominated convergence theorem we have
$$ mathfrakI = lim_r nearrow 1 I(r) , ,$$
where for $r in (0,1)$ we have defined
$$ I(r) = intlimits_0^1 mathrmd x,frac loglogfrac 1x(1+r x)^2 , . $$
With this regularisation interchanging summation and integration is actually justified and your calculations lead to
$$ I(r) = - gamma sum limits_n=0^infty (-r)^n - sum limits_n=0^infty (-r)^n log(1+n) equiv I_1 (r) + I_2(r) , . $$
The first sum is easy:
$$ I_1(r) = - fracgamma1+r , , $$
so $lim_r nearrow 1 I_1(r) = - fracgamma2$ .

For the second sum we can write
beginalign
I_2(r) &= frac1r sum_n=1^infty (-r)^n log(n) \
&= frac12r sum_k=1^infty [2 r^2k log(2k) - r^2k-1 log(2k-1) - r^2k+1 log(2k+1)] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r) log(2k+1) - frac1r (1-r) log(2k-1)right] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r)^2 log(2k+1)right] , .
endalign
The second term can be estimated by
beginalign
frac(1-r)^22r sum_k=1^infty r^2k log(2k+1) &leq frac(1-r)^22r^2 sum_n=1^infty sqrtn r^n \
&= frac(1-r)^22r^2 operatornameLi_-1/2 (r) \
&= fracsqrtpi4 r^2 sqrt1-r + mathcalO left((1-r)^3/2right)
endalign
as $r nearrow 1$. The asymptotic behaviour of the polylogarithm can be deduced from the series given here (the second one below 2.). Now we can use the monotone convergence theorem and Wallis' product to find
$$ lim_r nearrow 1 I_2 (r) = frac12 sum_k=1^infty logleft(frac4k^24k^2-1right) = frac12 log left(prod_k=1^infty frac4k^24k^2-1right) = frac12 log left(fracpi2right) , . $$

Therefore
$$ mathfrakI = frac12 left[log left(fracpi2right) - gammaright]$$
as claimed.

share|cite|improve this answer

answered Jul 16 at 21:32

ComplexYetTrivial

2,702624

By the dominated convergence theorem we have
$$ mathfrakI = lim_r nearrow 1 I(r) , ,$$
where for $r in (0,1)$ we have defined
$$ I(r) = intlimits_0^1 mathrmd x,frac loglogfrac 1x(1+r x)^2 , . $$
With this regularisation interchanging summation and integration is actually justified and your calculations lead to
$$ I(r) = - gamma sum limits_n=0^infty (-r)^n - sum limits_n=0^infty (-r)^n log(1+n) equiv I_1 (r) + I_2(r) , . $$
The first sum is easy:
$$ I_1(r) = - fracgamma1+r , , $$
so $lim_r nearrow 1 I_1(r) = - fracgamma2$ .

For the second sum we can write
beginalign
I_2(r) &= frac1r sum_n=1^infty (-r)^n log(n) \
&= frac12r sum_k=1^infty [2 r^2k log(2k) - r^2k-1 log(2k-1) - r^2k+1 log(2k+1)] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r) log(2k+1) - frac1r (1-r) log(2k-1)right] \
&= frac12r sum_k=1^infty r^2k left[logleft(frac4k^24k^2-1right) + (1-r)^2 log(2k+1)right] , .
endalign
The second term can be estimated by
beginalign
frac(1-r)^22r sum_k=1^infty r^2k log(2k+1) &leq frac(1-r)^22r^2 sum_n=1^infty sqrtn r^n \
&= frac(1-r)^22r^2 operatornameLi_-1/2 (r) \
&= fracsqrtpi4 r^2 sqrt1-r + mathcalO left((1-r)^3/2right)
endalign
as $r nearrow 1$. The asymptotic behaviour of the polylogarithm can be deduced from the series given here (the second one below 2.). Now we can use the monotone convergence theorem and Wallis' product to find
$$ lim_r nearrow 1 I_2 (r) = frac12 sum_k=1^infty logleft(frac4k^24k^2-1right) = frac12 log left(prod_k=1^infty frac4k^24k^2-1right) = frac12 log left(fracpi2right) , . $$

Therefore
$$ mathfrakI = frac12 left[log left(fracpi2right) - gammaright]$$
as claimed.

share|cite|improve this answer

answered Jul 16 at 21:32

ComplexYetTrivial

2,702624

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 21:32

ComplexYetTrivial

2,702624

answered Jul 16 at 21:32

ComplexYetTrivial

2,702624

answered Jul 16 at 21:32

ComplexYetTrivial

2,702624

2,702624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by Mathematica: $frac12 left(log left(fracpi 2right)-gamma right)$
  – David G. Stork
  Jul 16 at 21:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by Mathematica: $frac12 left(log left(fracpi 2right)-gamma right)$
  – David G. Stork
  Jul 16 at 21:51
  

  
  
  
  

Confirmed by Mathematica: $frac12 left(log left(fracpi 2right)-gamma right)$
– David G. Stork
Jul 16 at 21:51

Confirmed by Mathematica: $frac12 left(log left(fracpi 2right)-gamma right)$
– David G. Stork
Jul 16 at 21:51

add a comment | 

 

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