A closed form for ∫01ln⁡(−ln⁡x)li2xxdx

taoigas

taoigas

Answered

2022-01-22

A closed form for 01ln(lnx)li2xxdx

Answer & Explanation

stamptsk

stamptsk

Expert

2022-01-23Added 23 answers

I will start from Vladimirs
rakije2v

rakije2v

Expert

2022-01-24Added 12 answers

Change the integration variable z=lnx. The integral becomes
I={}0ln(z)Ei2zdz,
where Eiz is the exponential integral:
Eiz={z}ettdt
In this form the integral can be evaluated by Mathematica:

I=π26ln222(1+γ)ln2,

where γ is the Euler-Mascheroni Constant.

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