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?
On
Let $a\geq 0$ and $$f(a)=\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)\ln x}x^a dx$$ You're really looking for the value of $f(0)$.
Note that $$\begin{split} f^\prime(a) &= \int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)}x^a dx\\ &= \int_0^1 (x^{m-1}-x^{n-1})\left( \sum_{k\geq 0} (-1)^k x^{kp}\right)x^a dx\\ &= \sum_{k\geq 0} (-1)^k\int_0^1 (x^{kp+a+m-1} -x^{kp+a+n-1}) dx\\ &= \sum_{k\geq 0} (-1)^k\left( \frac 1 {kp+a+m}-\frac 1 {kp+a+n}\right) \end{split}$$ Integrating back w.r.t. $a$, and noting that $\lim_{a\rightarrow+\infty}f(a)=0$ gives $$\begin{split} f(a) &= \sum_{k\geq 0} (-1)^k \ln\left( \frac{kp + a + m}{kp + a + n}\right)\\ &=\ln\left ( \frac{a+m}{a+n} \cdot \frac{p+a+n}{p+a+m}\cdot \frac{2p+a+m}{2p+a+n}\dots\right) \end{split} $$ Evaluating at $a=0$: $$\int_0^1\frac{x^{m-1}-x^{n-1}}{(1+x^p)\ln x} dx= \ln\left ( \frac{m}{n} \cdot \frac{p+n}{p+m}\cdot \frac{2p+m}{2p+n}\dots\right)$$
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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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I am going to evaluate the integral by differentiating its partner $$ \begin{aligned} I(a) &=\int_0^1 \frac{x^a-x^{n-1}}{\left(1+x^p\right) \ln x} dx\\ I^{\prime}(a) &=\int_0^1 \frac{x^a}{1+x^p} d x \\ &=\sum_{k=0}^{\infty} \frac{(-1)^k}{a+pk+1} \end{aligned} $$ Integrating back gives our integral $$ \begin{aligned} I &=I(m-1)-I(n-1) \\ &=\int_{n-1}^{m-1} \sum_{k=0}^{\infty} \frac{(-1)^k}{a+p k+1} d a \\ &=\sum_{k=0}^{\infty}(-1)^k \ln \left(\frac{m+p k}{n+p k}\right)\\&= \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right) \end{aligned} $$
\begin{align} &\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln x}\,dx \\ =&\int_0^1 \frac{x^{m -1}-x^{n -1}}{\ln x}\sum_{k\ge 0} (-x^p)^k \,dx\\ =&\ \sum_{k\ge 0}(-1)^k \int_0^1 \frac{x^{m +pk-1}-x^{n +pk -1}}{\ln x} \> \ \overset{u=-\ln x}{dx}\\ =&\ \sum_{k\ge 0} (-1)^k \int_0^\infty \frac{e^{-(n +pk)u}-e^{-(m +pk)u}}{u}du\\ =&\ \sum_{k\ge 0} (-1)^k\ln \frac{m+kp}{n+kp} = \ln \frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\cdots \end{align}