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This integral is a generalization of another integral which arose during calculations related to a prior question of mine. Specifically, by using this integral representation for polylogarithms, $$\mathrm{Li}_n(z)=-\frac{1}{(n-2)!}\int_0^1\frac{\ln(1-tz)}{t}\ln^{n-2}(1/t)\,dt,$$ it can be shown that $$\sum_{k=1}^\infty\int_0^e(-1)^{k-1}\mathrm{Li}_{2k}(\ln{x})\,dx=\int_0^1 e^{1/t}\frac{\cos(\ln{t})\mathrm{Ei}(1-1/t)}{t}dt-e\int_0^1\frac{\cos(\ln{t})\ln(1-t)}{t}dt.$$ This equivalence holds numerically, so I believe that the above link's restriction of $\mathrm{Li}_s(z)$ to $|z|\lt 1$ should actually be extended to $z\lt1$. The first integral seems nigh impossible, even without the inclusion of $\cos(\ln{t})$.

Digressions aside, this brings me to my conjecture: $$-\int_0^1 x^a\cos(\ln{x})\ln(1-x)\,dx\stackrel{?}{=}\frac{a+1+i}{2a^2+4a+4}H_{a+1-i}+\frac{a+1-i}{2a^2+4a+4}H_{a+1+i},\tag{1}$$ where $H_n$ represents the $n$-th harmonic number and $a\geq-1$.

I don't know much about the analytical properties of the harmonic numbers, and I know even less about their extension into the complex plane. Therefore, I'm lost on how to proceed.

I'm not sure if it'll help much, but I have solved a similar integral by using integration by parts to obtain a rational zeta series. Maybe I'm missing something, but this approach doesn't seem to lead anywhere for my conjecture above. $$-\int_0^1\frac{\cos(\ln{x})\ln(1-x^b)}{x}dx=\frac{\pi}{2}\coth\left(\frac{\pi}{b}\right)-\frac{b}{2};\:b\gt0.\tag{2}$$It's easy to show that $\frac12(\pi\coth(\pi)-1)=\frac i2(H_{-i}-H_i)$.

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teadawg1337
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1 Answers1

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Hint. One may perform the change of variable $x=e^{-u}$, $dx=-e^{-u}du$, $\ln x=-u$, then make an integration by parts to recognize a classic integral representation of the digamma function: $$ \psi(z)=-\gamma+\int_0^\infty \frac{e^{-t}-e^{-zt}}{1-e^{-t}}\,dt,\quad \text{Re}\:z>-1. \tag1 $$ One obtains $$ \begin{align} &-\int_0^1 x^a\cos(\ln{x})\ln(1-x)\,dx \\&=-\int_0^\infty e^{-au}\cos(u)\ln(1-e^{-u})\,e^{-u}du \\&=-\:\text{Re}\int_0^\infty e^{-(a+i+1)u}\ln(1-e^{-u})\,du \\&=-\:\text{Re}\left(\left[\frac{e^{-(a+i+1)u}-1}{-(a+i+1)} \cdot \ln(1-e^{-u})\right]_0^\infty-\frac{1}{(a+i+1)}\int_0^\infty \frac{\left(1-e^{-(a+i+1)u}\right)e^{-u}}{1-e^{-u}}\,du\right) \\&=\frac{(a+1)\gamma}{(a+1)^2+1}+\text{Re}\:\frac{\psi(a+2+i)}{a+1+i} \end{align} $$ which is equivalent to the given expression in the conjecture.

Edit. One may use $$ \psi(z)=-\gamma+\int_0^1 \frac{1-u^{z-1}}{1-u}\,du,\quad \text{Re}\:z>0, \tag2 $$ which equivalent to $(1)$, avoiding the above change of variable.

Olivier Oloa
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