Consider the following integral: I=\int_0^\infty \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}dx

kramtus51 2022-01-06 Answered
Consider the following integral:
I=0x12x1ln(2x1)dx
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Expert Answer

Ronnie Schechter
Answered 2022-01-07 Author has 27 answers
Sub u=log(2x1). Then x=log(1+eu)log2, dx=(1log2)(du1+eu. The integral then becomes
1log2du1+eulog(1+eu)log21u=12log22
ducosh(u2)log(1+eu)log2u
=12log220ducosh(u2)log(1+eu)log2u+12log220ducosh(u2)log(1+eu)log2u
The nasty pieces of the integral cancel, and we are left with
12log220ducosh(u2)=π2log22
as correctly conjured.
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enhebrevz
Answered 2022-01-08 Author has 25 answers
I=0x12x1ln(2x1)dx
With the change of variables z2x1x=ln(1+z)ln(2), I is reduced to
I=1ln2(2)0ln(1+z)ln(2)z12(1+z)ln(z)dz
Now, we split the integral from (0,1) and from (1,). In the second one, we makes the change z1z such that we are left with an integration over (0,1):
I=1ln2(2)01ln(1+z)ln(2)z12(1+z)ln(z)dz+1ln2(2)01ln(1+1z)ln(2)z12(1+1z)[ln(z)]dzz2
=1ln2(2)01ln(1+z)ln(2)z12(1+z)ln(z)dz1ln2(2)
01ln(1+z)ln(z)ln(2)z12(1+z)ln(z)dz
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karton
Answered 2022-01-11 Author has 439 answers

Substitute (2x1)=t2 to get,
I=1ln220(ln(t2+1)ln2(t2+1)lnt)dt
Substitute t1t
I=1ln220(ln(t2+1)ln2(t2+1)lntdt+2ln220dtt2+1I=I+πln22I=π2ln22

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