Please verify this alternate proof of the Basel problem

Question

We know that:

$$\arctan(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+\cdots$$

Now,$$\int x^1\arctan(x) \,dx=\dfrac{x^3}{1\cdot3}-\dfrac{x^5}{3\cdot5}+\dfrac{x^7}{5\cdot7}+\cdots--(1)$$

$$\int x^3\arctan(x) \,dx=\dfrac{x^5}{1\cdot5}-\dfrac{x^7}{3\cdot7}+\dfrac{x^9}{5\cdot9}+\cdots--(2)$$

$$\int x^5\arctan(x) \,dx=\dfrac{x^7}{1\cdot7}-\dfrac{x^9}{3\cdot9}+\dfrac{x^11}{5\cdot11}+\cdots--(3)\cdots$$

Notice that if one were to add all these equations and replace $x$ with $i$, they would get the residual value from the square of the Leibniz pi formula times $i$.... I.E:

$$\dfrac{\pi}{4}\cdot\dfrac{\pi}{4}=\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots \right)\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots \right)$$

$$\dfrac{\pi^2}{16}=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\cdots+2\left(-\dfrac{1}{1\cdot3}+\dfrac{1}{1\cdot5}-\dfrac{1}{1\cdot7}+\cdots-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot7}+\cdots \right)$$

$$\dfrac{\pi^2}{16}=\left(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\cdots \right)+2K(Say)--(@)$$

Upon replacing $x=i$ in each of the mentioned equations$(1,2,3)$, it becomes apparent that:$$\boxed{iK=\sum_{n=0}^{\infty}\int_{0}^{i} I^{2n+1} dx=\sum_{n=0}^{\infty} \int_{0}^{i}x^{2n+1}\arctan(x) dx}$$, Where $I^{1}=(1); I^{3}=(2); I^{5}=(3)\cdots$

$$I^{1}=\int_{0}^{i} x^1\arctan(x) \,dx=\dfrac{(x^2+1)\arctan(x)-x}{2}$$ $$I^{3}=\int_{0}^{i} x^3\arctan(x) \,dx=\dfrac{(3x^4-3)\arctan(x)-x^3+3x}{12}$$ $$I^{5}=\int_{0}^{i} x^5\arctan(x) \,dx=\dfrac{(15x^6+15)\arctan(x)-3x^5+5x^3-15x}{90}$$

We get: $I^{1}=-\dfrac{i}{2}; I^{3}=\dfrac{i}{3}; I^{5}=-\dfrac{23i}{90}; I^{7}=\dfrac{22i}{105}\cdots$

Upon adding these values we get:

$$\boxed{iK=-\dfrac{i}{2}+\dfrac{i}{3}-\dfrac{23i}{90}+\dfrac{22i}{105}-\cdots--(*)}$$

Now, we know the Taylor series expansion for $\arctan^2(x)=x^2 - \dfrac{2}{3}x^4 + \dfrac{23}{45}x^6 - \dfrac{44}{105}x^8 + \dfrac{563}{1575}x^{10} + \cdots$

Now this is where I have doubts regarding equating two infinite series:

By observation, we can see that (*) is equal to the negative half of the $\arctan^2(x)$ expansion at $x=1$

Therefore, $$\dfrac{\arctan^2(1)}{2}=-(*)=-K=\dfrac{\pi^2}{32}=\left(\dfrac{1}{1\cdot3}-\dfrac{1}{1\cdot5}+\dfrac{1}{1\cdot7}-\cdots-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot7}-\cdots \right)$$

This result can be substituted in $(@)$, to get: $$\dfrac{\pi^2}{16}=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\cdots+2K=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\cdots-\dfrac{\pi^2}{16}$$

Which results in,$$\boxed{\dfrac{\pi^2}{8}=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+\dfrac{1}{11^2}+\dfrac{1}{13^2}+\dfrac{1}{15^2}+\cdots--(!)}$$

Now consider:$$L=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\cdots=(!)+(L/4)$$

$$=>\dfrac{3L}{4}=(!)=\dfrac{\pi^2}{8}$$

Therefore,$$\boxed{L=\dfrac{\pi^2}{6}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\cdots}$$

I haven't checked for convergence of any of the derived series and I have misused several infinite sequences mentioned here. Also I do not know if the integration equations hold for imaginary values of $x$. I would greatly appreciate if anyone could rectify this proof :)

Answer 1

Awesome observation! Rectifying your computation, let

$$ \chi_2(x) = \int_{0}^{x} \frac{\operatorname{artanh} t}{t} \, \mathrm{d}t = \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)^2} = x + \frac{x^3}{3^2} + \frac{x^5}{5^2} + \frac{x^7}{7^2} + \cdots $$

be the Legendre chi function of order $2$. Then OP's computation essentially boils down to verifying:

Lemma. For $0 \leq x \leq 1$, we have

\begin{align*} (\arctan x)^2 &= \chi_2(x^2) - \int_{0}^{x} \frac{2t}{1+t^2} \operatorname{artanh}(xt) \, \mathrm{d}t \tag{1} \\ &= \int_{0}^{x} \frac{2t}{1+t^2} \operatorname{artanh}(t/x) \, \mathrm{d}t. \tag{2} \end{align*}

Once this has been established, plugging $x = 1$ gives

$$ \frac{\pi^2}{16} = \chi_2(1) - \int_{0}^{1} \frac{2t}{1+t^2} \operatorname{artanh} t \, \mathrm{d}t = \int_{0}^{1} \frac{2t}{1+t^2} \operatorname{artanh} t \, \mathrm{d}t, $$

yielding $\chi_2(1) = \frac{\pi^2}{8}$ and hence the Basel's identity $\zeta(2) = \frac{\pi^2}{6}$.

---

So, it remains to prove $\text{(1)}$ and $\text{(2)}$. We will closely follow OP's computation (but in a rigorous manner) in order to establish both identites.

Proof of $(1)$. For $|x| < 1$, expanding the integrand as a power series in $t$ and applying Fubini's theorem gives

\begin{align*} -\int_{0}^{x} \frac{2t}{1+t^2} \operatorname{artanh}(xt) \, \mathrm{d}t &= -2 \int_{0}^{x} \biggl( \sum_{j=0}^{\infty} (-1)^j t^{2j+1} \biggr) \biggl( \sum_{k=0}^{\infty} \frac{(xt)^{2k+1}}{2k+1} \biggr) \, \mathrm{d}t \\ &= -2 \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} (-1)^j \frac{x^{2k+1}}{2k+1} \int_{0}^{x} t^{2j+2k+2} \, \mathrm{d}t \\ &= 2 \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1} \cdot \frac{(-1)^{k+j+1}x^{2k+2j+3}}{2k+2j+3} \\ &= 2 \sum_{0\leq k < l} \frac{(-1)^k x^{2k+1}}{2k+1} \cdot \frac{(-1)^{l}x^{2l+1}}{2l+1} \\ &= \biggl( \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1} \biggr)^2 - \sum_{k=0}^{\infty} \frac{x^{2(2k+1)}}{(2k+1)^2} \\ &= (\arctan x)^2 - \chi_2(x^2). \end{align*}

This proves $\text{(1)}$ for $|x| < 1$. Since both sides of $\text{(1)}$ are continuous at $x = 1$ from the left, the validity of $\text{(1)}$ extends to $x = 1$, proving the desired identity.

Proof of $(2)$. Let $|x| < 1$. Proceeding similarly as above,

\begin{align*} \int_{0}^{x} \frac{2t}{1+t^2} \operatorname{artanh}(t/x) \, \mathrm{d}t &= \int_{0}^{1} \frac{2x^2t}{1+x^2t^2} \operatorname{artanh} t \, \mathrm{d}t \\ &= 2 \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^j}{2k+1} x^{2j+2} \int_{0}^{1} t^{2k+2j+2} \, \mathrm{d}t \\ &= 2 \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^j x^{2j+2}}{(2k+1)(2k+2j+3)} \\ &= \sum_{j=0}^{\infty} (-1)^j x^{2j+2} \biggl( \sum_{k=0}^{\infty} \frac{2}{(2k+1)(2k+2j+3)} \biggr) \end{align*}

Now we focus on the inner sum in the last line. Applying partial fraction decomposition,

\begin{align*} \sum_{k=0}^{\infty} \frac{2}{(2k+1)(2k+2j+3)} &= \frac{2}{2j+2} \sum_{k=0}^{\infty} \biggl( \frac{1}{2k+1} - \frac{1}{2k+2j+3} \biggr) \\ &= \frac{2}{2j+2} \sum_{k=0}^{j} \frac{1}{2k+1} \\ &= \frac{1}{2j+2} \biggl(\sum_{k=0}^{j} \frac{1}{2k+1} + \sum_{k=0}^{j} \frac{1}{2(j-k)+1} \biggr) \\ &= \sum_{k=0}^{j} \frac{1}{(2k+1)(2(j-k)+1)}. \end{align*}

Plugging this back,

\begin{align*} \int_{0}^{x} \frac{2t}{1+t^2} \operatorname{artanh}(t/x) \, \mathrm{d}t &= \sum_{j=0}^{\infty} \sum_{k=0}^{j} \frac{(-1)^k x^{2k+1}}{2k+1} \cdot \frac{(-1)^{j-k} x^{2(j-k)+1}}{2(j-k)+1}, \end{align*}

which is the Cauchy product of the power series of $\arctan x$ with itself, completing the proof.

Answer 2

A solution 100% series free.

$x>0$, \begin{align}F(x)&=\int_0^x\frac{2t\text{arctanh}\left(\frac{t}{x}\right)}{1+t^2}dt\\ &\overset{u=\frac{1-\frac{t}{x}}{1+\frac{t}{x}}}=-2\int_0^1 \frac{(1-u)x^2\ln u}{(1+u)(x^2(1-u)^2+(1+u)^2)}du\\ &\overset{\text{IBP}}=\int_0^1 \left(\ln(1+u)-\frac{\ln\Big((1+u^2)(1+x^2)+2(1-x^2)u\Big)}{2}+\frac{\ln(1+x^2)}{2}\right)\frac{1}{u}du\\ &=\int_0^1\left(\frac{\ln(1+u)}{u}-\frac{\ln\left(1+u^2+2u\left(\frac{1-x^2}{1+x^2}\right)\right)}{2u}\right)du\\ \end{align}

$F$ can be extended continuously to $x=0$ and, using latter expression for $F$, one obtains $F(0)=0$.

\begin{align}F(x)-F(0)&=\int_0^x \frac{d}{dt}F(t)dt\\ &=\int_0^x \left(\int_0^1\frac{4t}{(1+t^2)\Big(t^2(1-u)^2+(1+u)^2\Big)}du\right)dt\\ &=2\int_0^x \left[\frac{\arctan\left(\frac{u(1+t^2)+(1-t^2)}{2t}\right)}{1+t^2}\right]_{u=0}^{u=1}dt\\ &=2\int_0^x \left(\frac{\arctan\left(\frac{1}{t}\right)-\arctan\left(\frac{1-t^2}{2t}\right)}{1+t^2}\right)dt\\ \end{align} Since $t>0$, \begin{align}\arctan\left(\frac{1}{t}\right)&=\frac{\pi}{2}-\arctan t\\ \arctan\left(\frac{2t}{1-t^2}\right)&=2\arctan t& \end{align} Therefore, \begin{align}F(x)-F(0)&=\int_0^x \frac{\left(\frac{\pi}{2}-\arctan t\right)-\left(\frac{\pi}{2}-\arctan\left(\frac{2t}{1-t^2}\right)\right)}{1+t^2}dt\\ &=2\int_0^x\frac{\arctan t}{1+t^2}dt=\left(\arctan x\right)^2\\ \end{align} Thus, \begin{align}\boxed{x\geq 0,F(x)=(\arctan x)^2}\end{align}

$0\leq x\leq 1$, \begin{align}W(x)&=\underbrace{\int_0^{x^2}\frac{\text{arctanh} t}{t}dt}_{u=\frac{t}{x}}-\int_0^{x}\frac{2t\text{arctanh}(tx)}{1+t^2}dt\\ &=\int_0^x\frac{\text{arctanh}(ux)}{u}du-\int_0^{x}\frac{2t\text{arctanh}(tx)}{1+t^2}dt\\ &=\int_0^{x}\frac{(1-t^2)\text{arctanh}(tx)}{t(1+t^2)}dt\\ &=\int_0^{x}\left(\int_0^1\frac{x(1-t^2)}{(1+t^2)(1-t^2u^2x^2)}du\right)dt\\\ &=\int_0^1 \left[\frac{2ux^2\text{arctanh}(tux)}{1+u^2x^2}-\frac{\text{arctanh}(tux)}{u}+\frac{2x\arctan t}{1+u^2x^2}\right]_{t=0}^{t=x}du\\ &=\underbrace{\int_0^1 \frac{2ux^2\text{arctanh}(ux^2)}{1+u^2x^2}du}_{z=ux}-\underbrace{\int_0^1 \frac{\text{arctanh}(ux^2)}{u}du}_{z=ux^2}+2(\arctan x)^2\\ &=\int_0^x\frac{2z\text{arctanh}(zx)}{1+z^2}dz-\int_0^{x^2}\frac{\text{arctanh}(z)}{z}dz+2(\arctan x)^2\\ &=-W(x)+2(\arctan x)^2\\ \end{align} Therefore, \begin{align}\boxed{0\leq x\leq 1,(\arctan x)^2=\int_0^{x^2}\frac{\text{arctanh} t}{t}dt-\int_0^{x}\frac{2t\text{arctanh}(tx)}{1+t^2}dt}\end{align}