The value of ∫01 log⁡sin⁡πx2dx is equal to

Let, I=∫01 log⁡sin⁡πx2dxput, πx2=t⇒dx=2πdtI=2π∫0π/2 log⁡sin⁡tdt=2πI1-----iwhere,I1=∫0π/2 log⁡sin⁡tdt=∫0π/2 log⁡sin⁡π2−tdt=∫0π/2 log⁡cos⁡tdt∴ 2I1=∫0π/2 (log⁡sin⁡t+log⁡cos⁡t)dt=∫0π/2 log⁡(sin⁡tcos⁡t)dt=∫0π/2 log⁡sin⁡2t2dt=∫0π/2 (log⁡sin⁡2t−log⁡2)dt=12∫0π log⁡sin⁡zdz−log⁡2∫0π/2 dt=12⋅2∫0π/2 log⁡sin⁡zdz−π2log⁡2⇒ 2I1=I1−π2log⁡2⇒I1=−π2log⁡2From Eq. (i), I=2π−π2log⁡2=−log⁡2