Prove that: ∫01xln⁡(1+x)1+x2dx=π296+ln228

Question

Prove that:  ∫01xln⁡(1+x)1+x2dx=π296+ln228

Foreckije · Accepted Answer

Putf(s)=∫01xln⁡(s+x)1+x2dxWe want to determine f(1). After differentiation we havef′(s)=∫01x(s+x)(1+x2)dx=−∫01s(s2+1)(s+x)dx+∫01sx+1(s2+1)(1+x2)dx=−sln⁡(1+s)1+s2+ln⁡s1+s2+sln⁡22(1+s2)+π4(1+s2)Now we integrate wrt s between 0 and 1. That yieldsf(1)−f(0)=−f(1)+f(0)+ln2(2)4+π216Consequentlyf(1)=f(0)+ln2(2)8+π232Butf(0)=∫01xln⁡(x)1+x2dx=∑k=0∞∫01(−1)kx2k+1ln⁡xdx=∑k=0∞(−1)k+14(k+1)2=−π248Finally we substitute that into (1)f(1)=ln2(2)8+π296

Serita Dewitt · Accepted Answer

This is a solution using complex analysis.  The integral  ∫01ln⁡x1+xdx=∑k=0∞(−1)k∫01xkln⁡xdx=∑k=1∞(−1)k1k2 (1)  =−π212  will be useful.  Put  I=∫01xln⁡(1+x)1+x2dx  After integration by parts we have  I=12ln22−12I1 (2)  where  I1=∫01ln⁡(1+x2)1+xdx  To evaluate I1 we integrate log⁡(1+z2)1+z over γ1+γ2, where γ1 is the imaginary axis from 0 to i and γ2 the unit circle from i to 1 clockwise. With log⁡z we mean ln|z|+iarg z, where −π&amp;lt;arg z&amp;lt;π  Put  I2=∫γ1log⁡(1+z2)1+zdz=∫01ln⁡(1−t2)1+itidt  =∫01ln⁡(1−t2)1+t2i(1−it)dt  Since I1 is real we are only interested in the real part of I2. After the substitution t=1s+1 we get  I2=∫01tln⁡(1−t2)1+t

nick1337 · Accepted Answer

An &#039;&#039;elementary&#039;&#039; solution. J=∫01xln⁡(1+x)1+x2dx, K=∫01xln⁡(1−x)1+x2dxJ+K=∫01xln⁡(1−x2)1+x2dx=12∫01ln⁡(2y1+y)1+ydy=12∫01ln⁡21+ydy+12∫01ln⁡y1+ydy−12∫01ln⁡(1+y)1+y=14ln2⁡2−124π2K−J=∫01xln⁡(1−x1+x)1+x2dx=∫01(1−y)ln⁡y(1+y)(1+y2)dy=∫01(ln⁡y1+y−yln⁡y1+y2)dy=−π212−∫01yln⁡y1+y2dy=−π212−14∫01ln⁡z1+zdz=−116π2Therefore,J=12((J+K)−(K−J))=12(14ln2⁡2−124π2+116π2)=18ln2⁡2+196π2

Prove that: \int_0^1\frac{x\ln(1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}

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