We have published the correct result for the integral listed in Bierens de Haan, Equation (1) in Table 148. I am attaching the Notebook with some plots and link to our paper
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What is the question?
I’d like to note that with a little handholding, Mathematica can be made to spit out this closed form.
Split the integral as
\int_0^1 \frac{\log\log t}{1+t^2} \mathrm dt+\int_1^\infty \frac{\log\log t}{1+t^2} \mathrm dt
The first readily evaluates:
f1 = Integrate[Log[Log[t]]/(1 + t^2), {t, 0, 1}]
1/4 π (I π + Log[(4 π^3)/Gamma[1/4]^4])
The second one is a little recalcitrant, but it succumbs to the substitution t\mapsto1/u:
f2 = Integrate[Log[Log[1/u]]/(1 + u^2), {u, 0, 1}]
1/4 π Log[(4 π^3)/Gamma[1/4]^4]
It should be noted at this juncture that f2 is also the real part of f1. Adding these two results up should give an equivalent result:
π/4 (I π + 2 Log[(4 π^3)/Gamma[1/4]^4]) ==
π/4 (I π + Log[4 π^2 (Gamma[3/4]/Gamma[1/4])^4]) // FullSimplify
True
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Why the second one you have to substitute t → 1/u? Thank you.
Here is another integral from Bierens de Haan Equation (13) Table 147
Because the second integral remains unevaluated without the substitution. Did you try it out yourself?
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Hi Mariusz, I am just posting integrals as an idea that I had. I am just looking for comments or other methods of deriving such integrals.
Where I find this Bierens de Haan Table 148?
Edited:
Ok I found this Book.
Regards M.I.
Yes, and the second one is the integrate function from 0 to Infinity or from 0 to 1 as your f2? I confused a little bit with this. And if replace t->1/u, should the denominator of f2 become 1+(1/u)^2? Thank you.
What do you think should be the expression for \mathrm dt in terms of \mathrm du?
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