<s>[INST] Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)$$
How does one prove this step-by-step?
 [/INST] First, we start with the following.
$$I(m,1)=\int_0^1\frac{x^{m-1}}{1+x}dx$$
We can substitute $u=-\ln{x}$ to get the following integral.
$$I(m,1)=\int_0^\infty\frac{e^{-mu}}{1+e^{-u}}du$$
When we apply the geometric series to $\frac{1}{1-(-e^{-u})}$, switch the order of integration and summation, then integrate, we get the following sum. We evaluate it by splitting it into even and odd terms and applying the digamma function.
$$I(m,1)=\sum_{n=0}^\infty\frac{(-1)^n}{n+m}=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+m-1}=\sum_{n=1}^\infty\left(\frac{(-1)^{(2n-1)-1}}{(2n-1)+m-1}-\frac{(-1)^{(2n)-1}}{(2n)+m-1}\right)$$
$$=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{1}{n+\frac{m-2}{2}}-\frac{1}{n+\frac{m-1}{2}}\right)=\frac{1}{2}\psi\left(\frac{m+1}{2}\right)-\frac{1}{2}\psi\left(\frac{m}{2}\right)$$
Next, we have to expand $I(m)$ with a new parameter.
$$I(m,p)=\int_0^1\frac{x^m}{1+x^p}dx$$
First, we can substitute $u=x^p$ to get the following
$$I(m,p)=\frac{1}{p}\int_0^1\frac{x^{\frac{m+1}{p}-1}}{1+x}dx=\frac{1}{2p}\psi\left(\frac{\frac{m+1}{p}+1}{2}\right)-\frac{1}{2p}\psi\left(\frac{\frac{m+1}{p}}{2}\right)$$
$$=\frac{1}{2p}\psi\left(\frac{m+p+1}{2p}\right)-\frac{1}{2p}\psi\left(\frac{m+1}{2p}\right)$$
Therefore
$$\int_0^1\frac{x^m}{1+x^p}dx=\frac{1}{2p}\psi\left(\frac{m+p+1}{2p}\right)-\frac{1}{2p}\psi\left(\frac{m+1}{2p}\right)$$
We take the definite integral with respect to m from 0 to a new parameter $n$
$$\int_0^1\frac{1}{1+x^p}\left(\int_0^n x^mdm\right)dx=\int_0^n\left(\frac{1}{2p}\psi\left(\frac{m+p+1}{2p}\right)-\frac{1}{2p}\psi\left(\frac{m+1}{2p}\right)\right)dm$$
$$\int_0^1\frac{1}{1+x^p}\frac{x^n-1}{\ln{x}}dx=\ln\left(\frac{\Gamma\left(\frac{m+p+1}{2p}\right)}{\Gamma\left(\frac{m+1}{2p}\right)}\right)\Bigg|_{m=0}^{m=n}=\ln\left(\frac{\Gamma\left(\frac{n+p+1}{2p}\right)\Gamma\left(\frac{1}{2p}\right)}{\Gamma\left(\frac{p+1}{2p}\right)\Gamma\left(\frac{n+1}{2p}\right)}\right)$$
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