I am trying to prove that

∫10log(x)log(1−x4)1+x2dx=π316−3Glog(2)

where G is Catalan’s Constant.

I was able to express it in terms of Euler Sums but it does not seem to be of any use.

∫10log(x)log(1−x4)1+x2dx= 116∞∑n=1ψ1(1/4+n)−ψ1(3/4+n)n

Here ψn(z) denotes the polygamma function.

Can you help me solve this problem ?.

**Answer**

I tried substitutions and the differentiation w.r.t a paramater trick like the other posters. Another partial result, or a trail of breadcrumbs to follow, is the following. We try a series expansion,

log(1−x4)1+x2=∞∑k=1x4k(x2−1)Hk,

where Hk are the Harmonic numbers. Then

∫10logxlog(1−x4)1+x2 dx=∞∑k=1Hk∫10x4k(x2−1)logx dx=∞∑k=1Hk(4k+1)2−∞∑k=1Hk(4k+3)2.

These sums look very similar to the ones evaluated in this post, in which they are transformed into alternating sums. Using the same techniques, or perhaps working back from the answers, we can hopefully show that

∞∑k=1Hk(4k+1)2=−G(π4+log82)+74ζ(3)+π332−π216log8,

∞∑k=1Hk(4k+3)2=−G(π4−log82)+74ζ(3)−π332−π216log8,

Subtracting the second from the first gives us

π316−Glog8.

**Attribution***Source : Link , Question Author : Shobhit Bhatnagar , Answer Author : Community*