Question

Evaluate the integral: 32i7π∫0∞xln2⁡(x)(ln⁡(ln⁡(x2)))x4−1dx

Answer 1

Let I=∫0∞xln2⁡(x)ln(ln⁡(x2))x4−1dx. Step 1: Substitution Set t=ln⁡(x2)⇒x=et/2,dx=12et/2dt. Then ln⁡x=t2,xx4−1dx=12ete2t−1dt=14dtsinh⁡t. Thus, I=116∫−∞∞t2ln⁡tsinh⁡tdt. Step 2: Symmetry and branch cut Split the integral: ∫−∞∞=∫0∞+∫−∞0. Using ln⁡(−t)=ln⁡t+iπ,sinh⁡(−t)=−sinh⁡t, we obtain cancellation of the real parts and are left with I=−iπ16∫0∞t2sinh⁡tdt. Step 3: Apply the prefactor 32i7πI=32i7π(−iπ16∫0∞t2sinh⁡tdt)=27∫0∞t2sinh⁡tdt. Step 4: Gamma--Zeta evaluation Use the expansion 1sinh⁡t=2∑n=0∞e−(2n+1)t. Then ∫0∞t2sinh⁡tdt=2∑n=0∞∫0∞t2e−(2n+1)tdt. Using ∫0∞t2e−atdt=2a3, we get ∫0∞t2sinh⁡tdt=4∑n=0∞1(2n+1)3=4(1−123)ζ(3)=72ζ(3).

Therefore, 32i7π∫0∞xln2⁡(x)ln⁡(ln⁡(x2))x4−1dx=ζ(3)≈1.202056903159594