How to prove, by real methods that
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$
where $H_n$ is the harmonic number and $\zeta$ is the Riemann zeta function.
This alternating Euler sum was already evaluated by M.N.C.E here using complex analysis and also by Cornel using series manipulation. My question here is can we do it by integration only?
The integral representation of the sum is $\ \frac16\int_0^1\frac{\ln^3x\ln(1+x)}{x(1+x)}dx$.
Thanks.
Starting with the generating function:
$$\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$$
replace $x$ with $-x$ then multiply both sides by $-\frac{\ln^3x}{6x}$ and integrate from $x=0$ to $1$ and use the fact that $-\frac16\int_0^1 x^{n-1}\ln^3xdx=\frac1{n^4}$ we get
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac16\int_0^1\frac{\ln^3x\ln(1+x)}{x(1+x)}dx=\frac16\mathcal{I}\tag1$$
$$\mathcal{I}=\int_0^1\frac{\ln^3x\ln(1+x)}{x(1+x)}dx=\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx-\underbrace{\int_1^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx}_{x\mapsto 1/x}$$
$$\mathcal{I}=\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx+\color{blue}{\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx}-\int_0^1\frac{\ln^4x}{1+x}dx$$
By adding $\ \mathcal{I}=\int_0^1\frac{\ln^3x\ln(1+x)}{x(1+x)}dx=\int_0^1\frac{\ln^3x\ln(1+x)}{x}dx-\color{blue}{\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx}\ $ to both sides, the blue integral nicely cancels out and we get
$$2\mathcal{I}=\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx-\int_0^1\frac{\ln^4x}{1+x}dx+\underbrace{\int_0^1\frac{\ln^3x\ln(1+x)}{x}dx}_{IBP}$$
$$2\mathcal{I}=\underbrace{\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx}_{\text{Beta function:}\ 6\zeta(2)\zeta(3)+6\zeta(5)}-\frac54\underbrace{\int_0^1\frac{\ln^4x}{1+x}dx}_{\frac{45}2\zeta(5)}$$
or
$$\mathcal{I}=3\zeta(2)\zeta(3)-\frac{177}{16}\zeta(5)\tag2$$
By plugging $(2)$ in $(1)$ we obtain that
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$
Bonus:
Following the same strategy, we can come up with two generalizations:
$$i)\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx=\frac12\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx+\frac{1+2a}2(2a-1)!\operatorname{Li}_{2a+1}(-1)$$
where the $\int_0^\infty$ integral can be done by beta function and $\operatorname{Li}_{a}(-1)=(2^{1-a}-1)\zeta(a)$.
Proof can be found here if needed.
$$ii)\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x)}{x(1+x)}dx=(2a)!(2a+2)\left(1-2^{-2a-1}\right)\zeta(2a+2)$$
Its interesting to get the result of $ii$ without using beta function.
Edit
Details for evaluating $\displaystyle\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}\ dx$ using beta function can be found here. Another way is to start with subbing $\frac{1}{1+x}\mapsto x$
$$\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}\ dx=\int_0^1\frac{\ln^3\left(\frac{x}{1-x}\right)\ln x}{1-x}\ dx$$
$$=\int_0^1\frac{\ln^4x}{1-x}-3\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}+3\underbrace{\int_0^1\frac{\ln^2x\ln^2(1-x)}{1-x}}_{IBP}-\underbrace{\int_0^1\frac{\ln x\ln^3(1-x)}{1-x}\ dx}_{IBP}$$
$$=\int_0^1\frac{\ln^4x}{1-x}-3\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}+2\underbrace{\int_0^1\frac{\ln^3(1-x)\ln x}{x}}_{\large 1-x\to x}-\frac14\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{\large 1-x\to x}$$
$$=\frac34\int_0^1\frac{\ln^4x}{1-x}\ dx-\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx$$ $$=\frac34(4!\zeta(5))+\sum_{n=1}^\infty H_n\int_0^1 x^n \ln^3x\ dx$$
$$=18\zeta(5)-6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}$$
$$=18\zeta(5)-6\sum_{n=1}^\infty\frac{H_n}{n^4}+6\zeta(5)$$
$$=18\zeta(5)-6[3\zeta(5)-\zeta(2)\zeta(3)]+6\zeta(5)$$
$$=6\zeta(2)\zeta(3)+6\zeta(5)$$
A first way may be found in the article A new powerful strategy of calculating a class of alternating Euler sums by Cornel Ioan Valean, which presents a very simple way of calculating the general case, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^{2m}}$.
A second way by Cornel of getting the result involves the use of a special form of the celebrated Beta function, $\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} \textrm{d}x = \operatorname{B}(a,b)$, as seen in the last part here https://math.stackexchange.com/q/3531956.
The extraction of the series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}$ is achieved immediately by noting that
$$\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^3 \partial b}\operatorname{B}(a,b)-\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^2 \partial b^2}\operatorname{B}(a,b)$$ $$=\underbrace{\int_0^1 \frac{\log^2(x)\log^2(1+x)}{x}\textrm{d}x}_{\displaystyle 15/4\zeta(5)-4\sum_{n=1}^{\infty} (-1)^{n-1} H_n/n^4}-\underbrace{\int_0^1 \frac{\log^3(x)\log(1+x)}{x}\textrm{d}x}_{\displaystyle -45/8 \zeta(5)}.$$
To avoid calculating two Beta function limits, one can be easily expressed in terms of the other, and all gets reduced to a classical Euler sum.
A third way of getting the value of the series is based on the identity $$\int_0^{\infty } \tanh (\pi x)\left(\frac{1}{x}-\frac{x}{n^2+x^2}\right) \textrm{d}x=2 H_{2n}-H_n,$$ which is presented in (Almost) Impossible Integrals, Sums, and Series. We have to follow a very similar strategy to the one presented in the first part of the solution here https://math.stackexchange.com/q/3495138 and we're done (theoretically we can calculate infinitely many series of this type, and far more advance with no touch of complex numbers, like $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^{4}}$, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{4n}}{n^{4}}$, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{8n}}{n^{6}}$, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{16n}}{n^{6}}$, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{32n}}{n^{8}}$).
A note: Very soon a new preprint dealing with this series will appear.
