Can this integral be solved with contour integral or by some application of residue theorem? ∫0∞log⁡(1+x)1+x2dx=π4log⁡2+ Catalan constant

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Can this integral be solved with contour integral or by some application of residue theorem?  ∫0∞log⁡(1+x)1+x2dx=π4log⁡2+  Catalan constant

eninsala06 · Accepted Answer

∫0∞log⁡(x+1)x2+1dx=∫01log⁡(x+1)x2+1dx+∫1∞log⁡(x+1)x2+1dx  =∫01log⁡(x+1)x2+1dx+∫01log⁡(x−1+1)x2+1dx  =2∫01log⁡(x+1)x2+1dx−∫01log⁡xx2+1dx  For the first integral, we plug  u=1−x1+x, dx=−2(u+1)2du  Then it is easy to find that  ∫01log⁡(x+1)x2+1dx=∫01log⁡2−log⁡(u+1)u2+1du=π4log⁡2−∫01log⁡(u+1)u2+1du  and hence  ∫01log⁡(x+1)x2+1dx=π8log⁡2  For the second integral, we plug x=e−t and we have  ∫01log⁡

Elaine Verrett · Accepted Answer

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karton · Accepted Answer

In this answer, the substitution x=1−y1+y is used to get ∫01log⁡(1+x)1+x2dx=π8log⁡(2) We can use the substitution x↦1/x to get ∫1∞log⁡(1+x)1+x2dx=∫01log⁡(1+x)−log⁡(x)1+x2dx which implies ∫0∞log⁡(1+x)1+x2dx=2∫01log⁡(1+x)1+x2dx−∫01log⁡(x)1+x2dx Therefore, we can use ∫01xklog⁡(x)dx=1k+1∫01log⁡(x)dxk+1=−1k+1∫01xk+1dlog⁡(x)=−1k+1∫01xkdx=−1(k+1)2 to get ∫0∞log⁡(1+x)1+x2=2∫01log⁡(1+x)1+x2dx−∫01log⁡(x)1+x2dx =π4log⁡(2)−∫01∑k=0∞(−1)kx2klog⁡(x)dx =π4log⁡(2)+∑k=0∞(−1)k(2k+1)2 =π4log⁡(2)+G

Can this integral be solved with contour integral or by

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