I am trying to evaluate the following integral, but Integrate just returns the input:
Integrate[2 E^(2 t) ExpIntegralEi[-2 t] Log[E^(-t)/t], {t, 0, x}]
Is it possible to find the analytic solution of this integral with Mathematica?
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2 Answers
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Another way:
Integrate[
2 E^(2 t) ExpIntegralEi[-2 t] Log[E^(-t)/t] // PowerExpand,
{t, 0, x}, Assumptions -> x > 0]
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Assumptions -> x > t > 0makes no logical sense, it should beAssumptions -> x > 0. $\endgroup$ Commented Oct 27 at 16:18 -
$\begingroup$ Good trick with
PowerExpandWill remove my answer using numerical integration as Mathematica can do it analytically. $\endgroup$– NasserCommented Oct 27 at 16:20 -
2$\begingroup$ @azerbajdzan You may find sometimes that
x > t > 0performs better than justx > 0. It may not make logical sense to you, but I have observed it to make a difference in what the internal code produces. And I have observed adding it does not always yield the simplification it should. Further, the edge cases change with each version update, naturally. It may be eventually in some version, it won't make a difference; but I think it should, since the path from0toxmay be complex, even ifxis real. I pass these remarks along as a trick one might try whenx > 0is insufficient. $\endgroup$ Commented Oct 27 at 16:35 -
$\begingroup$ @MichaelE2 Do you have an example when
x > 0fails butx > t > 0is successful? $\endgroup$ Commented Oct 27 at 17:41 -
1$\begingroup$ @azerbajdzan Nope. Specific examples have not stuck in my memory and such edge cases have not come up very often. There may be some on site among the 750+ answers I've given that involve Integrate. Further, what made a difference in, say, V8 no longer made a difference perhaps in V10 or later.
x > t > 0is just a way to make available the assumption that the path of integration is real. The person who might know is Daniel Lichtblau. Maybe it has become part of the internal code, and over the next few years I will realize that it hasn't helped. Or not. $\endgroup$ Commented Oct 27 at 18:00
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Mathematica is able to evaluate it for small integers x=1, x=2, ...
By substituting such small integers for x you can guess the formula of the integral for general x.
int=(-(EulerGamma*(1 + 2*EulerGamma)) + 2*x -
8*x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, 2*x] -
2*Log[2]^2 - Log[4]*Log[x] - Log[2*x] -
E^(2*x)*ExpIntegralEi[-2*x]*(-1 + 2*x + Log[x^2]) +
ExpIntegralEi[2*x]*(2*EulerGamma + Log[4*x^2]) -
EulerGamma*Log[16*x^2] +
Derivative[2, 0, 0][Hypergeometric1F1][0, 1, 0] -
Derivative[2, 0, 0][Hypergeometric1F1][0, 1, 2*x])/2
Which can be verified by numeric integration.
NIntegrate[
2 E^(2 t) ExpIntegralEi[-2 t] Log[E^-t/t], {t, 0, Pi + Sqrt[2]}]
N[int /. x -> Pi + Sqrt[2]]
1.69802
1.69802
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$\begingroup$ It should be noted that your formula is valid only for $x>0$. $\endgroup$– DomenCommented Oct 27 at 16:11
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1$\begingroup$ I do not expect OP wants
x<0as he/she surely can confirm. $\endgroup$ Commented Oct 27 at 16:13
e^(2t)should beExp[2*t]You start with indefinite integration, then you see it can't solve it. No need to start with definite integration. It is because you haveExp[2*t]in front. If you remove this term, then it can do it. i.sstatic.net/9QiM5CGK.png $\endgroup$