A tricky limit involving exponential integrals The Next CEO of Stack OverflowAn exponential improper integralProblem with evaluating the first term of Integration by PartsLimit of integral (involving sine integral)Calculate exponential limit involving trigonometric functionsEvaluate a limit involving definite integralLimit of these integralsClean Limit ProofLimit in a fraction of integrals involving an DEEvaluating $lim_nrightarrowinfty int_0^pi frac sin x1+ cos^2 (nx) dx$Integrals involving multi-valued functions
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A tricky limit involving exponential integrals
The Next CEO of Stack OverflowAn exponential improper integralProblem with evaluating the first term of Integration by PartsLimit of integral (involving sine integral)Calculate exponential limit involving trigonometric functionsEvaluate a limit involving definite integralLimit of these integralsClean Limit ProofLimit in a fraction of integrals involving an DEEvaluating $lim_nrightarrowinfty int_0^pi frac sin x1+ cos^2 (nx) dx$Integrals involving multi-valued functions
$begingroup$
We define exponential integral according to https://en.wikipedia.org/wiki/Exponential_integral#Definition_by_Ein
as
$$textEi_n(x) = int_1^infty frace^-xtt^n dt$$
I'm trying to evaluate the limit of the sequence
$$ -fraci ln(2) lim_nrightarrow infty left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] $$
Where $n$ is evaluated over positive integers.
This approaches real number which is very close to the value $frac1pi$ but slightly less than it. (I have confirmed this experimentally).
Work so far:
Attempting to evaluate this is as a limit of a function has been futile,
Since my inner limit results is an expression that doesn't converge:
$$lim_nrightarrow infty, nin mathbbN left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] = lim_n rightarrow infty, n in mathbbN int_1^inftyfrace^-2^-ntt^1 - frac2ipiln2 dt - int_1^inftyfrace^-2^-ntt^1 + frac2ipiln2dt $$
= $$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty frace^-2^-nt left( t^1 + frac2ipiln2 - t^1 - frac2ipiln2 right)t^2 dt right]$$
$$ =lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
At this point, "letting $n$ go to infinity" yields the divergent integral:
$$ lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty fracsin frac2piln 2 ln tt dt right] $$
Some Developments:
On Suggestion of Sangchul Lee I tried to apply integration by parts to our second to last term (dropping the $2i$ for now).
$$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
$$ = lim_nrightarrow infty, nin mathbbN left[ -fracln 22pi e^-2^-nt cos left( frac2piln 2 ln t right)_@[1,infty] - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
We can now simplify the first term (taking the evaluation of infinity)
$$ = lim_nrightarrow infty, nin mathbbN left[ fracln 22pi e^-2^-n - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And we see this becomes:
$$ = fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ 2^-n int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And, after consulting mathematica this reduces to:
$$ fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ frac12 left( Gamma[1 - frac2ipiln 2 , 2^-n ] + Gamma[1 + frac2ipiln 2 , 2^-n ] right)right] $$
And that reduces pretty simply to:
$$ fracln 22pi - fracln 22 pi left[ frac12 left( Gamma[1 - frac2ipiln 2 , 0 ] + Gamma[1 + frac2ipiln 2 , 0 ] right)right] $$
And Wolfram is able to verify that this indeed is the result I expect (once I multiply by $frac-i(2i)ln 2$ ) as it is within $10^-6$ of $frac1pi$.
My question remains: Is there a better closed form for this term?
sequences-and-series limits definite-integrals improper-integrals lacunary-series
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
$endgroup$
add a comment |
$begingroup$
We define exponential integral according to https://en.wikipedia.org/wiki/Exponential_integral#Definition_by_Ein
as
$$textEi_n(x) = int_1^infty frace^-xtt^n dt$$
I'm trying to evaluate the limit of the sequence
$$ -fraci ln(2) lim_nrightarrow infty left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] $$
Where $n$ is evaluated over positive integers.
This approaches real number which is very close to the value $frac1pi$ but slightly less than it. (I have confirmed this experimentally).
Work so far:
Attempting to evaluate this is as a limit of a function has been futile,
Since my inner limit results is an expression that doesn't converge:
$$lim_nrightarrow infty, nin mathbbN left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] = lim_n rightarrow infty, n in mathbbN int_1^inftyfrace^-2^-ntt^1 - frac2ipiln2 dt - int_1^inftyfrace^-2^-ntt^1 + frac2ipiln2dt $$
= $$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty frace^-2^-nt left( t^1 + frac2ipiln2 - t^1 - frac2ipiln2 right)t^2 dt right]$$
$$ =lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
At this point, "letting $n$ go to infinity" yields the divergent integral:
$$ lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty fracsin frac2piln 2 ln tt dt right] $$
Some Developments:
On Suggestion of Sangchul Lee I tried to apply integration by parts to our second to last term (dropping the $2i$ for now).
$$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
$$ = lim_nrightarrow infty, nin mathbbN left[ -fracln 22pi e^-2^-nt cos left( frac2piln 2 ln t right)_@[1,infty] - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
We can now simplify the first term (taking the evaluation of infinity)
$$ = lim_nrightarrow infty, nin mathbbN left[ fracln 22pi e^-2^-n - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And we see this becomes:
$$ = fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ 2^-n int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And, after consulting mathematica this reduces to:
$$ fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ frac12 left( Gamma[1 - frac2ipiln 2 , 2^-n ] + Gamma[1 + frac2ipiln 2 , 2^-n ] right)right] $$
And that reduces pretty simply to:
$$ fracln 22pi - fracln 22 pi left[ frac12 left( Gamma[1 - frac2ipiln 2 , 0 ] + Gamma[1 + frac2ipiln 2 , 0 ] right)right] $$
And Wolfram is able to verify that this indeed is the result I expect (once I multiply by $frac-i(2i)ln 2$ ) as it is within $10^-6$ of $frac1pi$.
My question remains: Is there a better closed form for this term?
sequences-and-series limits definite-integrals improper-integrals lacunary-series
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
$endgroup$
1
$begingroup$
Hint: Take integration by parts to the integral right before taking limit.
$endgroup$
– Sangchul Lee
Mar 19 at 23:11
$begingroup$
@SangchulLee so this somehow worked and numerically converges to the right thing, although I was hoping for a more beautiful closed form. It seems mathematica can't turn the Gamma terms into something elementary either, is there any further simplifications that can be made here in your opinion?
$endgroup$
– frogeyedpeas
Mar 20 at 3:54
2
$begingroup$
I will try to add an answer when I finish my dinner cooking, but the point is that the limit will be a kind of 'averaged version' (a.k.a. regularization) of the integral of sine, and the usual way of justifying this is to perform integration by parts so as to obtain better regularity control of the integrand. (To put simple, oscillatory behavior is not good to deal with, so averaging it out is often a better way to go.)
$endgroup$
– Sangchul Lee
Mar 20 at 4:02
add a comment |
$begingroup$
We define exponential integral according to https://en.wikipedia.org/wiki/Exponential_integral#Definition_by_Ein
as
$$textEi_n(x) = int_1^infty frace^-xtt^n dt$$
I'm trying to evaluate the limit of the sequence
$$ -fraci ln(2) lim_nrightarrow infty left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] $$
Where $n$ is evaluated over positive integers.
This approaches real number which is very close to the value $frac1pi$ but slightly less than it. (I have confirmed this experimentally).
Work so far:
Attempting to evaluate this is as a limit of a function has been futile,
Since my inner limit results is an expression that doesn't converge:
$$lim_nrightarrow infty, nin mathbbN left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] = lim_n rightarrow infty, n in mathbbN int_1^inftyfrace^-2^-ntt^1 - frac2ipiln2 dt - int_1^inftyfrace^-2^-ntt^1 + frac2ipiln2dt $$
= $$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty frace^-2^-nt left( t^1 + frac2ipiln2 - t^1 - frac2ipiln2 right)t^2 dt right]$$
$$ =lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
At this point, "letting $n$ go to infinity" yields the divergent integral:
$$ lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty fracsin frac2piln 2 ln tt dt right] $$
Some Developments:
On Suggestion of Sangchul Lee I tried to apply integration by parts to our second to last term (dropping the $2i$ for now).
$$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
$$ = lim_nrightarrow infty, nin mathbbN left[ -fracln 22pi e^-2^-nt cos left( frac2piln 2 ln t right)_@[1,infty] - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
We can now simplify the first term (taking the evaluation of infinity)
$$ = lim_nrightarrow infty, nin mathbbN left[ fracln 22pi e^-2^-n - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And we see this becomes:
$$ = fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ 2^-n int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And, after consulting mathematica this reduces to:
$$ fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ frac12 left( Gamma[1 - frac2ipiln 2 , 2^-n ] + Gamma[1 + frac2ipiln 2 , 2^-n ] right)right] $$
And that reduces pretty simply to:
$$ fracln 22pi - fracln 22 pi left[ frac12 left( Gamma[1 - frac2ipiln 2 , 0 ] + Gamma[1 + frac2ipiln 2 , 0 ] right)right] $$
And Wolfram is able to verify that this indeed is the result I expect (once I multiply by $frac-i(2i)ln 2$ ) as it is within $10^-6$ of $frac1pi$.
My question remains: Is there a better closed form for this term?
sequences-and-series limits definite-integrals improper-integrals lacunary-series
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
$endgroup$
We define exponential integral according to https://en.wikipedia.org/wiki/Exponential_integral#Definition_by_Ein
as
$$textEi_n(x) = int_1^infty frace^-xtt^n dt$$
I'm trying to evaluate the limit of the sequence
$$ -fraci ln(2) lim_nrightarrow infty left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] $$
Where $n$ is evaluated over positive integers.
This approaches real number which is very close to the value $frac1pi$ but slightly less than it. (I have confirmed this experimentally).
Work so far:
Attempting to evaluate this is as a limit of a function has been futile,
Since my inner limit results is an expression that doesn't converge:
$$lim_nrightarrow infty, nin mathbbN left[ textEi_1- frac2ipiln(2) left( 2^-n right)-textEi_1+frac2ipiln(2) left( 2^-n right) right] = lim_n rightarrow infty, n in mathbbN int_1^inftyfrace^-2^-ntt^1 - frac2ipiln2 dt - int_1^inftyfrace^-2^-ntt^1 + frac2ipiln2dt $$
= $$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty frace^-2^-nt left( t^1 + frac2ipiln2 - t^1 - frac2ipiln2 right)t^2 dt right]$$
$$ =lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
At this point, "letting $n$ go to infinity" yields the divergent integral:
$$ lim_nrightarrow infty, nin mathbbN left[ 2i int_1^infty fracsin frac2piln 2 ln tt dt right] $$
Some Developments:
On Suggestion of Sangchul Lee I tried to apply integration by parts to our second to last term (dropping the $2i$ for now).
$$ lim_nrightarrow infty, nin mathbbN left[ int_1^infty e^-2^-n t fracsin frac2piln 2 ln tt dt right]$$
$$ = lim_nrightarrow infty, nin mathbbN left[ -fracln 22pi e^-2^-nt cos left( frac2piln 2 ln t right)_@[1,infty] - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
We can now simplify the first term (taking the evaluation of infinity)
$$ = lim_nrightarrow infty, nin mathbbN left[ fracln 22pi e^-2^-n - 2^-n fracln 22 pi int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And we see this becomes:
$$ = fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ 2^-n int_1^infty e^-2^-ntcos left( frac2piln 2 ln t right) dt right] $$
And, after consulting mathematica this reduces to:
$$ fracln 22pi - fracln 22 pi lim_nrightarrow infty, nin mathbbN left[ frac12 left( Gamma[1 - frac2ipiln 2 , 2^-n ] + Gamma[1 + frac2ipiln 2 , 2^-n ] right)right] $$
And that reduces pretty simply to:
$$ fracln 22pi - fracln 22 pi left[ frac12 left( Gamma[1 - frac2ipiln 2 , 0 ] + Gamma[1 + frac2ipiln 2 , 0 ] right)right] $$
And Wolfram is able to verify that this indeed is the result I expect (once I multiply by $frac-i(2i)ln 2$ ) as it is within $10^-6$ of $frac1pi$.
My question remains: Is there a better closed form for this term?
sequences-and-series limits definite-integrals improper-integrals lacunary-series
sequences-and-series limits definite-integrals improper-integrals lacunary-series
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
edited Mar 20 at 3:56
frogeyedpeas
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
asked Mar 19 at 22:06
frogeyedpeasfrogeyedpeas
7,67372054
7,67372054
1
$begingroup$
Hint: Take integration by parts to the integral right before taking limit.
$endgroup$
– Sangchul Lee
Mar 19 at 23:11
$begingroup$
@SangchulLee so this somehow worked and numerically converges to the right thing, although I was hoping for a more beautiful closed form. It seems mathematica can't turn the Gamma terms into something elementary either, is there any further simplifications that can be made here in your opinion?
$endgroup$
– frogeyedpeas
Mar 20 at 3:54
2
$begingroup$
I will try to add an answer when I finish my dinner cooking, but the point is that the limit will be a kind of 'averaged version' (a.k.a. regularization) of the integral of sine, and the usual way of justifying this is to perform integration by parts so as to obtain better regularity control of the integrand. (To put simple, oscillatory behavior is not good to deal with, so averaging it out is often a better way to go.)
$endgroup$
– Sangchul Lee
Mar 20 at 4:02
add a comment |
1
$begingroup$
Hint: Take integration by parts to the integral right before taking limit.
$endgroup$
– Sangchul Lee
Mar 19 at 23:11
$begingroup$
@SangchulLee so this somehow worked and numerically converges to the right thing, although I was hoping for a more beautiful closed form. It seems mathematica can't turn the Gamma terms into something elementary either, is there any further simplifications that can be made here in your opinion?
$endgroup$
– frogeyedpeas
Mar 20 at 3:54
2
$begingroup$
I will try to add an answer when I finish my dinner cooking, but the point is that the limit will be a kind of 'averaged version' (a.k.a. regularization) of the integral of sine, and the usual way of justifying this is to perform integration by parts so as to obtain better regularity control of the integrand. (To put simple, oscillatory behavior is not good to deal with, so averaging it out is often a better way to go.)
$endgroup$
– Sangchul Lee
Mar 20 at 4:02
1
1
$begingroup$
Hint: Take integration by parts to the integral right before taking limit.
$endgroup$
– Sangchul Lee
Mar 19 at 23:11
$begingroup$
Hint: Take integration by parts to the integral right before taking limit.
$endgroup$
– Sangchul Lee
Mar 19 at 23:11
$begingroup$
@SangchulLee so this somehow worked and numerically converges to the right thing, although I was hoping for a more beautiful closed form. It seems mathematica can't turn the Gamma terms into something elementary either, is there any further simplifications that can be made here in your opinion?
$endgroup$
– frogeyedpeas
Mar 20 at 3:54
$begingroup$
@SangchulLee so this somehow worked and numerically converges to the right thing, although I was hoping for a more beautiful closed form. It seems mathematica can't turn the Gamma terms into something elementary either, is there any further simplifications that can be made here in your opinion?
$endgroup$
– frogeyedpeas
Mar 20 at 3:54
2
2
$begingroup$
I will try to add an answer when I finish my dinner cooking, but the point is that the limit will be a kind of 'averaged version' (a.k.a. regularization) of the integral of sine, and the usual way of justifying this is to perform integration by parts so as to obtain better regularity control of the integrand. (To put simple, oscillatory behavior is not good to deal with, so averaging it out is often a better way to go.)
$endgroup$
– Sangchul Lee
Mar 20 at 4:02
$begingroup$
I will try to add an answer when I finish my dinner cooking, but the point is that the limit will be a kind of 'averaged version' (a.k.a. regularization) of the integral of sine, and the usual way of justifying this is to perform integration by parts so as to obtain better regularity control of the integrand. (To put simple, oscillatory behavior is not good to deal with, so averaging it out is often a better way to go.)
$endgroup$
– Sangchul Lee
Mar 20 at 4:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write $alpha = frac2pilog 2$ for simplicity. Then we are interested in the limit of the following quantity
beginalign*
I_n
:= fracoperatornameEi_1 - ialpha (2^-n) - operatornameEi_1 + ialpha (2^-n)ilog 2.
endalign*
Plugging the definition of $operatornameEi_s$ and taking integration by parts,
beginalign*
I_n
&= frac1ilog 2 int_1^infty fract^ialpha - t^-ialphate^-2^-nt , mathrmdt
= frac2log 2 int_1^infty fracsinleft( alpha log t right)te^-2^-nt , mathrmdt \
&hspace1em= frace^-2^-npi - frac1pi int_1^infty cosleft( alpha log t right) 2^-n e^-2^-nt , mathrmdt.
endalign*
Now apply the substitution $u = 2^-nt$ and notice that
$$cos(alpha log t) = cos(alphalog u + alpha nlog 2) = cos(alpha log u) $$
since $alpha log 2 = 2pi$. Then
beginalign*
lim_ntoinfty I_n
&= lim_ntoinfty left[ frace^-2^-npi - frac1pi int_2^-n^infty cosleft( alpha log u right) e^-u , mathrmdu right] \
&= frac1pi - frac1pi int_0^infty cosleft( alpha log u right) e^-u , mathrmdu
endalign*
By recalling that the gamma function is defined as $Gamma(s) = int_0^infty u^s-1e^-u , mathrmdu$, this reduces to
beginalign*
lim_ntoinfty I_n
&= frac1pi - fracGamma(1+ialpha) + Gamma(1-ialpha)2pi
= frac1pioperatornameReleft[ 1 - Gamma(1+ialpha) right].
endalign*
But I am skeptical of this having an elementary closed form.
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
add a comment |
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1 Answer
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oldest
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active
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active
oldest
votes
$begingroup$
Write $alpha = frac2pilog 2$ for simplicity. Then we are interested in the limit of the following quantity
beginalign*
I_n
:= fracoperatornameEi_1 - ialpha (2^-n) - operatornameEi_1 + ialpha (2^-n)ilog 2.
endalign*
Plugging the definition of $operatornameEi_s$ and taking integration by parts,
beginalign*
I_n
&= frac1ilog 2 int_1^infty fract^ialpha - t^-ialphate^-2^-nt , mathrmdt
= frac2log 2 int_1^infty fracsinleft( alpha log t right)te^-2^-nt , mathrmdt \
&hspace1em= frace^-2^-npi - frac1pi int_1^infty cosleft( alpha log t right) 2^-n e^-2^-nt , mathrmdt.
endalign*
Now apply the substitution $u = 2^-nt$ and notice that
$$cos(alpha log t) = cos(alphalog u + alpha nlog 2) = cos(alpha log u) $$
since $alpha log 2 = 2pi$. Then
beginalign*
lim_ntoinfty I_n
&= lim_ntoinfty left[ frace^-2^-npi - frac1pi int_2^-n^infty cosleft( alpha log u right) e^-u , mathrmdu right] \
&= frac1pi - frac1pi int_0^infty cosleft( alpha log u right) e^-u , mathrmdu
endalign*
By recalling that the gamma function is defined as $Gamma(s) = int_0^infty u^s-1e^-u , mathrmdu$, this reduces to
beginalign*
lim_ntoinfty I_n
&= frac1pi - fracGamma(1+ialpha) + Gamma(1-ialpha)2pi
= frac1pioperatornameReleft[ 1 - Gamma(1+ialpha) right].
endalign*
But I am skeptical of this having an elementary closed form.
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
add a comment |
$begingroup$
Write $alpha = frac2pilog 2$ for simplicity. Then we are interested in the limit of the following quantity
beginalign*
I_n
:= fracoperatornameEi_1 - ialpha (2^-n) - operatornameEi_1 + ialpha (2^-n)ilog 2.
endalign*
Plugging the definition of $operatornameEi_s$ and taking integration by parts,
beginalign*
I_n
&= frac1ilog 2 int_1^infty fract^ialpha - t^-ialphate^-2^-nt , mathrmdt
= frac2log 2 int_1^infty fracsinleft( alpha log t right)te^-2^-nt , mathrmdt \
&hspace1em= frace^-2^-npi - frac1pi int_1^infty cosleft( alpha log t right) 2^-n e^-2^-nt , mathrmdt.
endalign*
Now apply the substitution $u = 2^-nt$ and notice that
$$cos(alpha log t) = cos(alphalog u + alpha nlog 2) = cos(alpha log u) $$
since $alpha log 2 = 2pi$. Then
beginalign*
lim_ntoinfty I_n
&= lim_ntoinfty left[ frace^-2^-npi - frac1pi int_2^-n^infty cosleft( alpha log u right) e^-u , mathrmdu right] \
&= frac1pi - frac1pi int_0^infty cosleft( alpha log u right) e^-u , mathrmdu
endalign*
By recalling that the gamma function is defined as $Gamma(s) = int_0^infty u^s-1e^-u , mathrmdu$, this reduces to
beginalign*
lim_ntoinfty I_n
&= frac1pi - fracGamma(1+ialpha) + Gamma(1-ialpha)2pi
= frac1pioperatornameReleft[ 1 - Gamma(1+ialpha) right].
endalign*
But I am skeptical of this having an elementary closed form.
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
add a comment |
$begingroup$
Write $alpha = frac2pilog 2$ for simplicity. Then we are interested in the limit of the following quantity
beginalign*
I_n
:= fracoperatornameEi_1 - ialpha (2^-n) - operatornameEi_1 + ialpha (2^-n)ilog 2.
endalign*
Plugging the definition of $operatornameEi_s$ and taking integration by parts,
beginalign*
I_n
&= frac1ilog 2 int_1^infty fract^ialpha - t^-ialphate^-2^-nt , mathrmdt
= frac2log 2 int_1^infty fracsinleft( alpha log t right)te^-2^-nt , mathrmdt \
&hspace1em= frace^-2^-npi - frac1pi int_1^infty cosleft( alpha log t right) 2^-n e^-2^-nt , mathrmdt.
endalign*
Now apply the substitution $u = 2^-nt$ and notice that
$$cos(alpha log t) = cos(alphalog u + alpha nlog 2) = cos(alpha log u) $$
since $alpha log 2 = 2pi$. Then
beginalign*
lim_ntoinfty I_n
&= lim_ntoinfty left[ frace^-2^-npi - frac1pi int_2^-n^infty cosleft( alpha log u right) e^-u , mathrmdu right] \
&= frac1pi - frac1pi int_0^infty cosleft( alpha log u right) e^-u , mathrmdu
endalign*
By recalling that the gamma function is defined as $Gamma(s) = int_0^infty u^s-1e^-u , mathrmdu$, this reduces to
beginalign*
lim_ntoinfty I_n
&= frac1pi - fracGamma(1+ialpha) + Gamma(1-ialpha)2pi
= frac1pioperatornameReleft[ 1 - Gamma(1+ialpha) right].
endalign*
But I am skeptical of this having an elementary closed form.
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
Write $alpha = frac2pilog 2$ for simplicity. Then we are interested in the limit of the following quantity
beginalign*
I_n
:= fracoperatornameEi_1 - ialpha (2^-n) - operatornameEi_1 + ialpha (2^-n)ilog 2.
endalign*
Plugging the definition of $operatornameEi_s$ and taking integration by parts,
beginalign*
I_n
&= frac1ilog 2 int_1^infty fract^ialpha - t^-ialphate^-2^-nt , mathrmdt
= frac2log 2 int_1^infty fracsinleft( alpha log t right)te^-2^-nt , mathrmdt \
&hspace1em= frace^-2^-npi - frac1pi int_1^infty cosleft( alpha log t right) 2^-n e^-2^-nt , mathrmdt.
endalign*
Now apply the substitution $u = 2^-nt$ and notice that
$$cos(alpha log t) = cos(alphalog u + alpha nlog 2) = cos(alpha log u) $$
since $alpha log 2 = 2pi$. Then
beginalign*
lim_ntoinfty I_n
&= lim_ntoinfty left[ frace^-2^-npi - frac1pi int_2^-n^infty cosleft( alpha log u right) e^-u , mathrmdu right] \
&= frac1pi - frac1pi int_0^infty cosleft( alpha log u right) e^-u , mathrmdu
endalign*
By recalling that the gamma function is defined as $Gamma(s) = int_0^infty u^s-1e^-u , mathrmdu$, this reduces to
beginalign*
lim_ntoinfty I_n
&= frac1pi - fracGamma(1+ialpha) + Gamma(1-ialpha)2pi
= frac1pioperatornameReleft[ 1 - Gamma(1+ialpha) right].
endalign*
But I am skeptical of this having an elementary closed form.
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
answered Mar 20 at 20:32
Sangchul LeeSangchul Lee
96.3k12171282
96.3k12171282
add a comment |
add a comment |
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$begingroup$
Hint: Take integration by parts to the integral right before taking limit.
$endgroup$
– Sangchul Lee
Mar 19 at 23:11
$begingroup$
@SangchulLee so this somehow worked and numerically converges to the right thing, although I was hoping for a more beautiful closed form. It seems mathematica can't turn the Gamma terms into something elementary either, is there any further simplifications that can be made here in your opinion?
$endgroup$
– frogeyedpeas
Mar 20 at 3:54
2
$begingroup$
I will try to add an answer when I finish my dinner cooking, but the point is that the limit will be a kind of 'averaged version' (a.k.a. regularization) of the integral of sine, and the usual way of justifying this is to perform integration by parts so as to obtain better regularity control of the integrand. (To put simple, oscillatory behavior is not good to deal with, so averaging it out is often a better way to go.)
$endgroup$
– Sangchul Lee
Mar 20 at 4:02