Consider the following integral: I=∫0∞x−12x−1ln⁡(2x−1)dx

Question

Consider the following integral:  I=∫0∞x−12x−1ln⁡(2x−1)dx

Ronnie Schechter · Accepted Answer

Sub u=log⁡(2x−1). Then x=log⁡(1+eu)log⁡2, dx=(1log⁡2)(du1+e−u. The integral then becomes  1log⁡2∫−∞∞du1+e−ulog⁡(1+eu)log⁡2−1u=12log22  ∫−∞∞ducosh(u2)log⁡(1+eu)−log⁡2u  =12log22∫−∞0ducosh(u2)log⁡(1+eu)−log⁡2u+12log22∫0∞ducosh(u2)log⁡(1+eu)−log⁡2u  The nasty pieces of the integral cancel, and we are left with  12log22∫0∞ducosh(u2)=π2log22  as correctly conjured.

enhebrevz · Accepted Answer

I=∫0∞x−12x−1ln⁡(2x−1)dx  With the change of variables z≡2x−1→x=ln⁡(1+z)ln⁡(2), I is reduced to  I=1ln2(2)∫0∞ln⁡(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 z→1z 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)z−12(1+1z)[−ln⁡(z)]dzz2  =1ln2(2)∫01ln⁡(1+z)−ln⁡(2)z12(1+z)ln⁡(z)dz−1ln2(2)  ∫01ln⁡(1+z)−ln⁡(z)−ln⁡(2)z12(1+z)ln⁡(z)dz

karton · Accepted Answer

Substitute (2x−1)=t2 to get, I=1ln2⁡2∫0∞(ln⁡(t2+1)−ln⁡2(t2+1)ln⁡t)dt Substitute t↦1t →I=−1ln2⁡2∫0∞(ln⁡(t2+1)−ln⁡2(t2+1)ln⁡tdt+2ln2⁡2∫0∞dtt2+1→I=−I+πln2⁡2→I=π2ln2⁡2

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

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New Questions in Calculus 1