I'm currently trying to find a closed form for:
$$\int_{0}^1 \frac{u\ln(2)-\ln(1+u)}{\ln(u)u}du$$
I derived this using frullani's integral and the sum:
$$\sum_{n=1}^\infty \frac{\ln(n)(-1)^n}{n}$$
I've tried finding some differentiating under the integral, but no luck. I can't seem to connect it to the beta, gamma, or zeta function either.
Best Answer
$$\begin{align} \int_0^1 \frac{x\ln(2)-\ln(1+x)}{x\ln(x)}dx &=\int_{-\infty}^0 \frac{e^x\ln(2)-\ln(1+e^x)}{x}dx\\ &=\big[(e^x\ln(2)-\ln(1+e^x))\ln|x|\big]_{-\infty}^0-\int_{-\infty}^0 \bigg(e^x\ln(2)-\frac{e^x}{1+e^x}\bigg)\ln|x|dx\\ &=-\int_{-\infty}^0 \bigg(e^x\ln(2)-\frac{e^x}{1+e^x}\bigg)\ln|x|dx\\ &=\int_{0}^\infty \bigg(\frac{1}{1+e^x}-e^{-x}\ln(2)\bigg)\ln(x)dx\\ &=\int_{0}^\infty \frac{\ln(x)}{1+e^x}dx-\ln(2)\int_0^\infty e^{-x}\ln(x)dx \\ &=\int_{0}^\infty \frac{\ln(x)}{1+e^x}dx+\gamma\ln(2) \end{align}$$
Now we just need to calculate the integral $$\int_0^\infty \frac{\ln(x)}{1+e^x}dx$$ This has been done before on MSE, but I'll show it again here so you don't have to go hunting. $$\begin{align} \int_0^\infty \frac{\ln(x)}{1+e^x}dx &=\lim_{\epsilon\to 0}\bigg(\int_\epsilon^\infty \frac{\ln(x)}{1+e^x}dx\bigg)\\ &=\lim_{\epsilon\to 0}\bigg(\int_\epsilon^\infty \frac{\ln(x)}{e^x-1}dx-\int_\epsilon^\infty \frac{2\ln(x)}{e^{2x}-1}dx\bigg)\\ &=\lim_{\epsilon\to 0}\bigg(\int_\epsilon^\infty \frac{\ln(x)}{e^x-1}dx-\int_{2\epsilon}^\infty \frac{\ln(x/2)}{e^{x}-1}dx\bigg)\\ &=\lim_{\epsilon\to 0}\bigg(\int_\epsilon^{2\epsilon} \frac{\ln(x)}{e^x-1}dx+\int_{2\epsilon}^\infty \frac{\ln(2)}{e^{x}-1}dx\bigg)\\ &=\lim_{\epsilon\to 0}\bigg(\int_\epsilon^{2\epsilon} \frac{\ln(x)}{x}dx+\int_{2\epsilon}^\infty \frac{\ln(2)}{e^{x}-1}dx\bigg)\tag{1}\\ &=\lim_{\epsilon\to 0}\bigg(\ln^2(2\epsilon)/2-\ln^2(\epsilon)/2+\int_{2\epsilon}^\infty \frac{\ln(2)}{e^{x}-1}dx\bigg)\\ &=\lim_{\epsilon\to 0}\bigg(\ln^2(2)/2+\ln(2)\ln(\epsilon)+\int_{2\epsilon}^\infty \frac{\ln(2)}{e^{x}-1}dx\bigg)\\ &=\lim_{\epsilon\to 0}\bigg(\ln^2(2)/2+\ln(2)\ln(\epsilon)-\ln(2\epsilon)\ln(2)\bigg)\tag{2}\\ &=-\ln^2(2)/2 \end{align}$$ where equalities $(1)$ and $(2)$ are justified by the fact that $\frac{1}{e^x-1}\sim \frac{1}{x}$ as $x\to 0$. Thus, we have that your integral is equal to $$\int_0^1 \frac{x\ln(2)-\ln(1+x)}{x\ln(x)}dx=\gamma\ln(2)-\frac{\ln^2(2)}{2}$$