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Don't fear divergencies: $\int_0^{\infty}\frac{\tanh^3(x)}{x^2}\,dx$ ?

Posted 7 years ago

NOTE: the notebook with full content is attached at the end of the post.


On Mathoverflow, the question about a closed form for the integral $\int_0^{\infty}\frac{\tanh^3(x)}{x^2}\,dx$ was raised. Currently, Mathematica can't give a closed form answer for this integral.

Integrate[Tanh[x]^3/x^2, {x, 0, ?}]

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But we can easily get a numerical value.

num = NIntegrate[Tanh[x]^3/x^2, {x, 0, ?},  
WorkingPrecision -> 100, PrecisionGoal -> 30] // N[#, 30] &

1.15478531332317626405907045194

How could one fine a closed form result? With Integrate, Sum, SeriesCoefficient, ... we have a lot of tools at our hands to do some experimental mathematics. A simple manual change of variables does not help.

Integrate[y^3/(ArcTanh[y]^2)  Dt[ArcTanh[y], y], {y, 0, 1}]

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The next typical trick is expanding the integrand in a series and integrating term by term:

TrigToExp[Tanh[x]^3]

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SeriesCoefficient[% /. {Exp[-x] -> g, Exp[x] -> 1/g}, {g, 0, n}] // 
 FullSimplify[#, n ? Integers] &

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The resulting integrals do diverge:

Integrate[Exp[-n x]/x^2 , {x, 0, ?}]

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We can make the integral convergent by adding a factor. For the case $?=0$ that we need, the integral diverges:

Integrate[Exp[-n x]/x^2 x^? , {x, 0, ?}]

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And the sum can be done in closed form for general $?$:

Sum[n^(1 - ?) Gamma[-1 + ?] Cos[(n ?)/2] (2 + n^2), {n, ?}]

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Expanding the result around $?=0$ gives a finite leading term:

Series[%, {?, 0, 1}]

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And indeed this is the result we were looking for:

N[(-1 - (4 Log[2])/15 + 12 Log[Glaisher] - 120 Derivative[1][Zeta][-3]), 30]

1.15478531332317626405907045194

Alternatively, we can split of the divergent point from the integral:

Integrate[Exp[-n x]/x^2, {x, ?, ?}]

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The leading terms are:

Series[(E^(-n ?)/? - n Gamma[0, n ?]), {?, 0, 2}, Assumptions -> ? > 0]

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The n term does give no sum contribution:

Sum[n Cos[(n ?)/2] (2 + n^2), {n, ?}, Regularization -> "Dirichlet"]

0

And the $n \ln(n)$ term gives the same result as above:

Sum[n Log [n] Cos[(n ?)/2] (2 + n^2) n^?, {n, ?}]

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Series[%, {\[Alpha], 0, 1}]

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By conjecturing the constants involved in the integral, we could have used FindIntegerNullVector to get a conjecture for a closed form:

FindIntegerNullVector[{1.15478531332317626405907045194,1, 
Log[2], EulerGamma, Log[Glaisher], Derivative[1][Zeta][-3]}]

{-15, -15, -4, 0, 180, -1800}

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POSTED BY: Michael Trott
2 Replies
POSTED BY: Marvin Ray Burns

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