Introduction: At first I define: $$ \zeta(s)=\sum \limits_{n=1}^{\infty}\frac{1}{n^s}\quad\bigg\rvert\quad y=\sum\limits_{n=1}^{\infty}\frac{x^n}{n^s} \quad\bigg\rvert\quad y_{(n)}=\frac{d^ny}{dx^n} $$ In order to simplify the following, I also define a new operator: $$ x\frac{dy}{dx}=xD\implies x(\frac{d}{dx}(x(\frac{d}{dx}(...x\frac{dy}{dx})=xD^n $$ Using the new defined operator $s$ times on $y$, yields a geometric series: $$ xD^sy=\frac{x}{1-x} $$ By using the stirling numbers of the second kind $S(s,n)$ we can show the following equality: $$ S(s,n)=\frac{1}{n!}\sum\limits_{m=0}^{n}(-1)^{n-m}\binom{n}{m}m^s \\ xD^sy=\sum\limits_{n=1}^{s}S(s,n)x^ny_{(n)}=\frac{x}{1-x} $$ We can prove that this identity is indeed correct via induction over $s$:
Base case, $s=1$: $$ xDy=S(1,1)xy_{(1)}\implies True $$ In order to show the induction step, we need to use the recurrence relation between the stirling numbers of the second kind, which, similar to the binomial coefficients, states the following: $$ S(s+1,n)=nS(s,n)+S(s,n-1) $$ So for our Induction step, $s\implies s+1$, we get: $$ xD^{s+1}y=xD(xD^sy)=xD(\sum\limits_{n=1}^{s}S(s,n)x^ny_{(n)})=\sum\limits_{n=1}^{s}(S(s,n)nx^{n}y_{(n)}+S(s,n)x^{n+1}y_{n+1}) \\ =\sum\limits_{n=1}^{s}nS(s,n)x^ny_{(n)}+\sum\limits_{n=2}^{s+1}S(s,n-1)x^ny_{(n)} \\ =S(s,1)xy_{(1)}+\sum\limits_{n=2}^{s}S(s+1,n)x^ny_{(n)}+S(s,s)x^{s+1}y_{(s+1)} \\ =\sum\limits_{n=1}^{s+1}S(s+1,n)x^ny_{(n)}\implies True $$ Note that the last step is only possible because $S(s,1)=S(s+1,1)$ and $S(s,s)=1$.
Iam currently able to find a integral or a solution for some values of $s$.
Examples: Some of which are:
$s=1:$ $$ y_{(1)}=\frac{1}{1-x}\implies y=-\ln|1-x|+C $$ We can show that $C=0$, because: $$ y(0)=0\implies 0=-\ln|1|+C\implies C=0 $$ So if we evaluate $-\ln|1-x|$ at $x=1$, we can see that $\zeta(1)$ diverges: $$ \zeta(1)=\lim\limits_{\epsilon \searrow 0}\ln|\frac{1}{\epsilon}|=\infty $$ $s=2:$ $$ y_{(2)}+\frac{y_{(1)}}{x}=\frac{1}{x(1-x)} $$ Now substituting $u=y_{(1)}$, yealds: $$ u_{(1)}+\frac{u}{x}=\frac{1}{x(1-x)} $$ This can be easily solved using variation of constants: $$ u_{(1)}+A(x)u=B(x)\implies \alpha(x)=e^{\int A(x)dx}\implies u=\frac{1}{\alpha(x)}\int B(x)\alpha(x)dx+C \\ u=\frac{1}{x}\int \frac{1}{1-x}dx=-\frac{\ln|1-x|}{x}+C $$ Similar to the first example we can show that $C=0$, because: $$ u(0)=1+\lim\limits_{x\to 0}\sum\limits_{n=2}^{\infty}\frac{x^{n-1}}{n}=1 \\ 1=\lim\limits_{x \searrow 0}-\frac{\ln|1-x|}{x}+C\overset{LH}{=}\lim\limits_{x \searrow 0}\frac{1}{1-x}+C=1+C\implies C=0 $$ So we finally arrive at a new Integral, which evaluated at $x=1$ should give us the value of $\zeta(2)$: $$ \zeta(2)=\int\limits_{0}^{1}-\frac{\ln|1-x|}{x}dx=\frac{\pi ^2}{6} $$ Here iam only capable of showing this equality, because i already was aware of the vaule of $\zeta(2)$. The only way i were able to evaluate the given integral, was via the series expansion of $-\ln|1-x|$, which leads me right back to the beginning.
It was also possible for me to find a solution for $xD^3y=\frac{x}{1-x}$, using standard methods. $$ xD^3y=\frac{x}{1-x}\implies y_{(1)}+3xy_{(2)}+x^2y_{(3)}=\frac{1}{1-x} $$ Substituting $u=y_{(1)}$: $$ u+3xu_{(1)}+x^2u_{(2)}=\frac{1}{1-x} $$ Using $u=u_{comp}+u_{part}$, we get: $$ u+3xu_{(1)}+x^2u_{(2)}=0 $$ Letting $x=e^{w}$: $$ u_{(1)}=\frac{du}{dw}=\frac{du}{dx}\frac{dx}{dw}=\frac{du}{dx}e^u=\frac{du}{dx}x \\ u_{(2)}=\frac{d^2u}{dw^2}=\frac{d}{dx}(\frac{du}{dw})\frac{dx}{dw}=x^2\frac{d^2u}{dx^2}+x\frac{du}{dx} \\ u+2xu_{(1)}+(xu_{(1)}+x^2u_{(2)})=0\implies u+2\frac{du}{dw}+\frac{d^2u}{dw^2}=0 $$ So we get the characteristic polynomial: $$ 1+2\mu+\mu^2=0\implies \mu_{1/2}=-1 $$ Which gets us: $$ u(w)=e^{-w}(C_1+wC_2)\implies u_{comp}=\frac{1}{x}(C_1+\ln(x)C_2) $$ Now computing $u_{part}$:
At first we calculate the Wronskian of our summands of $u_{comp}$: $$ \mathcal{W}(x)=\begin{vmatrix} \frac{1}{x} & \frac{\ln(x)}{x} \\ -\frac{1}{x^2} & \frac{1-\ln(x)}{x^2} \end{vmatrix}=\frac{1}{x^3} $$ So via variation of parameters, we get: $$ u_{part}=\frac{\ln(x)}{x}\int\frac{1}{1-x}dx-\frac{1}{x}\int\frac{\ln(x)}{1-x}dx \\ =-\frac{\ln(x)\ln|1-x|}{x}+\frac{\ln(x)\ln|1-x|}{x}-\frac{1}{x}\int\frac{\ln|1-x|}{x}dx \\ =\frac{1}{x}\int-\frac{\ln|1-x|}{x}dx $$ Now combining $u_{comp}$ and $u_{part}$: $$ u=\frac{C_1}{x}+\frac{C_2\ln(x)}{x}+\frac{1}{x}\int-\frac{\ln|1-x|}{x}dx $$ Befor making the next step, we take a look at the given constant $C_1$: $$ u(1)=\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6} \\ \frac{\pi^2}{6}=C_1+\frac{\pi^2}{6}\implies C_1=0 $$ Because $u=y_{(1)}$, we get: $$ y=\int\frac{C_2\ln(x)}{x}+\frac{1}{x}\int-\frac{\ln|1-x|}{x}dxdx=\frac{1}{2}C_2\ln(x)\ln|x|+\int\frac{1}{x}\int-\frac{\ln|1-x|}{x}dxdx $$ Unfortunately i wasn't able to find a sufficient second boundary value, I tried $u(0)=1+\lim\limits_{x \to 0}\sum\limits_{n=2}^{\infty}\frac{x^{n-1}}{n^2}=1$ but if this is plugged into the solution for $u$ it diverges. Another attempt at showing that $C_2=0$ might be by the following idea: $$ \int\frac{1}{x}\int-\frac{\ln|1-x|}{x}dxdx=\sum\limits_{n=1}^{\infty}\frac{x^n}{n^3} \\ y=\frac{1}{2}C_2\ln(x)\ln|x|+\sum\limits_{n=1}^{\infty}\frac{x^n}{n^3} $$ Therefore, because of the definiton of $y$ it follows that $C_2=0$.
So we finally arrive at: $$ \zeta(3)=\int\limits_{0}^{1}\int\limits_{0}^{x}-\frac{\ln|1-t|}{xt}dtdx $$ Continuity: At this point you might asked yourself, why is it even possible to evaluate the given answers at $x=1$? Afterall the equality $\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$ only holds for $|x|<1$. I found an answer here "Evaluating the power series of arctan at x=1". The answer states that according to Abel's theorem, a function $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$ is continous on the intervall $(-1,1]$, if $\sum\limits_{n=0}^{\infty}a_n$ converges. That $\zeta(s)$ converges for $s>1$, has already been shown in this post "Convergence of $\zeta(s)$ on $\Re(s)>1$". For $\zeta(1)$ the series obvioulsy diverges, so does $\lim\limits_{x \nearrow 1}\frac{1}{1-x}$.
Question: How would one go about evaluating the given differential equations for lager values of $s$, maybe even in general? Also how would I be able to show that the presentet Integral $\int\limits_{0}^{1}-\frac{\ln|1-x|}{x}dx$ is indeed equal to $\frac{\pi^2}{6}$ without using series expansion?
