I'm trying to evaluate
$$\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx$$
I used the series expansions for $\ln(1-x)$ and managed to get a double summation out of them
$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+4m)}$$
Does anyone know how to go about computing this sum? I know that this has a closed form involving Apery's Constant and Catalan's constant but I'm unsure how to tackle this double summation.
Evaluate $\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx$
definite integralsintegrationsummation
Related Solutions
The definition of $\eta(s)$ is $$\eta(s)=(1-2^{1-s})\zeta(s)$$ Take the derivative, and we get by product rule $$\eta'(s)=2^{1-s}\ln(2)\zeta(s)+(1-2^{1-s})\zeta'(s)$$ $\eta(s)$ also has the following series representation $$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$$ Again, take the derivative $$\eta'(s)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^s}$$ Now, we can obtain a closed form $$\eta'(2)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}=2^{1-2}\ln(2)\zeta(2)+(1-2^{1-2})\zeta'(2)$$ Note that
$$\zeta'(2)=\frac{1}{6}\pi^2(-12\ln(A)+\ln(2\pi)+\gamma),~\zeta(2)=\frac{\pi^2}{6}$$
$$\begin{align} \sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}&=2^{-1}\ln(2)\zeta(2)+(1-2^{-1})\zeta'(2)\\ &=\frac{\pi^2}{12}\ln(2)+\frac{\pi^2}{12}(-12\ln(A)+\ln(2\pi)+\gamma)\\ &=\frac{\pi^2}{12}(\ln(4\pi)-12\ln(A)+\gamma) \end{align}$$
Set $$ \theta_2(q):=\sum^{\infty}_{n=-\infty}q^{(n+1/2)^2}\textrm{, }\theta_3(q):=\sum^{\infty}_{n=-\infty}q^{n^2}\textrm{, }\theta_4(q):=\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}. $$ Then $$ \theta_2(q)^2=\frac{2kK}{\pi}\textrm{, }\theta_3(q)^2=\frac{2K}{\pi}\textrm{, }\theta_4(q)^2=\frac{2k'K}{\pi} $$ and $$ \frac{dk}{dr}=-\frac{k(k')^2K(k)^2}{\pi\sqrt{r}}. $$ Thus $$ I=\int^{1}_{0}K\left(\sqrt{k}\right)^2dk=2\int^{1}_{0}K(k)^2kdk=-2\int^{0}_{\infty}K(k)^2k\frac{k(k')^2K(k)^2}{\pi\sqrt{r}}dr= $$ $$ =2\int^{\infty}_{0}\frac{(kk')^2K(k)^4}{\pi\sqrt{r}}dr=2\int^{\infty}_{0}\frac{\pi^2\theta_2(q)^4}{4K^2}\frac{\pi^2\theta_4(q)^4}{4K^2}\frac{K^4}{\pi\sqrt{r}}dr= $$ $$ =\frac{\pi^3}{8}\int^{\infty}_{0}\frac{\theta_2(q)^4\theta_4(q)^4}{\sqrt{r}}dr $$ But $q=e^{-\pi\sqrt{r}}$. Hence $dq=\frac{-\pi q}{2\sqrt{r}}dr$. Hence $$ I=\frac{-\pi^3}{8}\int^{0}_{1}\theta_2(q)^4\theta_4(q)^4\frac{1}{\sqrt{r}}\frac{2\sqrt{r}}{\pi q}dq=\frac{\pi^2}{4}\int^{1}_{0}\theta_2(q)^4\theta_4(q)^4\frac{dq}{q}. $$ From the above integral we conclude easily that $$ I=\frac{\pi^3}{2}\int^{\infty}_{0}\theta_2\left(e^{-2\pi t}\right)^4\theta_4\left(e^{-2\pi t}\right)^4dt. $$
We set now $$ P(z):=\theta_2(q)^4\theta_4(q)^4\textrm{, }q=e^{2\pi i z}\textrm{, }Im(z)>0 $$ The function $P(z)$ is a weight 4 modular form in $\Gamma_1(4)$. The space $M_4(\Gamma_1(4))$ has dimension 3, with no cusp forms i.e. $dim(S_4(\Gamma_1(4))=0$ and $dim(E_4(\Gamma_1(4))=3$.
Consider now the functions $$ E_{2k}(q):=2\zeta(2k)\left(1+\frac{2}{\zeta(1-2k)}\sum^{\infty}_{n=1}\sigma_{2k-1}(n)q^n\right), $$ where $\sigma_{\nu}(n)=\sum_{d|n}d^{\nu}$, $\zeta(s)$ being the Riemann zeta function. The functions $E_{2k}(q)$ are the classical Eisenstein series of weight $2k$, $k-$positive integer. For the present case we get $k=2$ and we will use the property $E_{2k}(q)-lE_{2k}(q^l)$ is a base element of $M_{2k}(\Gamma_1(N))$, when $l|N$.
Also in [1] I have proven that if $q=e^{2\pi i z}$, $Im(z)>0$, then $$ H_k(q):=\frac{\pi^k}{k!}\left(\left(2-2^k\right)|B_{k}|+4ki^kF_{k}(q)\right), $$ $$ F_k(q):=\sum^{\infty}_{n=1}\sigma^{*}_{k-1}(n)q^n, $$ where $\sigma^{*}_{\nu}(n):=\sum_{d|n,d-odd}d^{\nu}$, $B_{k}$ are the Bernoulli numbers, $k-$even positive integher, are modular forms of the space $M_k\left(\Gamma_1(2)\right)$, where $$ \Gamma_1(N):=\left\{\left[ \begin{array}{cc} a\textrm{ }b\\ c\textrm{ }d \end{array}\right]:a,b,c,d\in\textbf{Z}\textrm{, }ab-cd=1\textrm{, }a,d\equiv1(N)\textrm{ and }b,c\equiv 0(N) \right\}. $$ By this way and from the fact that $P(z)$ is in dimension 3 space, comparing coefficients, we have $$ P(z)=C_1\left(E_4(q)-4E_4(q^4)\right)+C_2H_4(q)+C_3H_4(-q), $$ where $$ C_1=-\frac{14}{5\pi^4}\textrm{, }C_2=\frac{28}{\pi^4}\textrm{, }C_3=-\frac{92}{5\pi^4}. $$
Hence writing $$ P(z)=a_P(0)+\sum^{\infty}_{n=1}a_P(n)q^n, $$ we get $$ a_P(0)=0 $$ and for $n=1,2,\ldots$, we get $$ a_P(n)=-\frac{224}{15}\sigma_3(n)+\frac{896}{15}\sigma_3\left(\frac{n}{4}\right)+\frac{56}{3}\sigma^{*}_3(n)-\frac{184}{15}(-1)^n\sigma^{*}_3(n) $$ The Dirichlet series $L(s)$ coresponding to $a_P(n)$ are $$ L(s)=\sum^{\infty}_{n=1}\frac{a_P(n)}{n^s} $$ and the function $$ \Lambda_P(s):=\left(\frac{2}{i}\right)^4\int^{+\infty}_{0}P(it)t^{s-1}dt=G(s)\left(\frac{2}{i}\right)^4\sum^{\infty}_{n=1}\frac{a_P(n)}{n^s}, $$ where $G(s)=(2\pi)^{-s}\Gamma(s)$ (here $\Gamma$ means the Euler's Gamma function), have the property (analytic continuation) via the functional equation $$ \Lambda_P(s)=4^{2-s}\Lambda_P(4-s) $$ Hence we want to find $\Lambda_P(1)=4\Lambda_P(3)$. But $$ \Lambda_P(s)=2^4(2\pi)^{-s}\Gamma(s)[-\frac{224}{15}\sum^{\infty}_{n=1}\frac{\sigma_3(n)}{n^s}+\frac{896}{15}4^{-s}\sum^{\infty}_{n=1}\frac{\sigma_3(n)}{n^{s}}+ $$ $$ +\frac{56}{3}\sum^{\infty}_{n=1}\frac{\sigma^{*}_3(n)}{n^s}-\frac{184}{15}\sum^{\infty}_{n=1}\frac{(-1)^n\sigma^{*}_3(n)}{n^s}]= $$ $$ =2^4(2\pi)^{-s} \Gamma(s)[-\frac{224}{15}\zeta(s-3)\zeta(s)+\frac{896}{15}4^{-s}\zeta(s-3)\zeta(s)+ $$ $$ +\frac{56}{3}2^{-s}(-8+2^s)\zeta(s-3)\zeta(s) -\frac{184}{15}2^{-s}\left(2^{1-s}-1\right)(-8+2^s)\zeta(s-3)\zeta(s)]. $$ Hence $$ \Lambda_P(3)=\lim_{s\rightarrow 3}\Lambda_P(s)=2^4 (2\pi)^{-3} \Gamma(3)7\zeta(3). $$ Hence $$ \Lambda_P(1)=4\Lambda_P(3)=\frac{28\zeta(3)}{\pi^3}=2\cdot 2^{4}I \pi^{-3} $$ and consequently $$ I=\frac{7\zeta(3)}{2}. $$ QED
REFERENCES
[1]: N.D. Bagis. ''Evaluations of certain theta functions in Ramanujan theory of alternative modular bases''. arXiv:1511.03716v2 [math.GM] 6 Dec 2017.
Best Answer
Utilize the referenced integrals below $$I_1= \int_0^1 \frac{\ln(1-x)\ln(1-x^2)}{x}dx=\frac{11}8\zeta(3)\>\>\>\>\>\>\>\>\>\>\>\>$$ $$I_2= \int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx=\frac{23}{32}\zeta(3)-\frac\pi2 G$$ to obtain
$$\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx= I_1+I_2=\frac{67 }{32}\zeta (3)-\frac{\pi }{2}G$$