Htykuut

So I was watching a video on YouTube about how $$sum_{i=1}^infty frac{chi(i)}{i} = frac{pi}{4}$$ (note that $chi(i) = 0$ for even numbers $i$, $1$ for $text{mod}(i, 4) = 1$, and $-1$ for $text{mod}(i,4) = 3$) and one of the proofs shown involved stating that $$sum_{i=1}^infty frac{chi(i)}{i} = int_{0}^{1} sum_{i=0}^{infty}chi(i+1)x^{i}dx,.$$
My question is 1.) how is this done and 2.) how can this be replicated with different infinite sums. Thanks in advance!

Note that, for all $igeq 1$,
$$frac{1}{i} = int_0^1 x^{i-1} dx$$
(this is a "trick" worth knowing), and therefore
$$
sum_{i=1}^infty frac{chi(i)}{i} =
sum_{i=1}^infty chi(i)int_0^1 x^{i-1} dx
=
int_0^1sum_{i=1}^infty chi(i)x^{i-1} dx
=
int_0^1sum_{k=0}^infty chi(k+1)x^{k} dx
$$
where the only part which would require justification is when we swap $int_0^1$ and $sum_{i=1}^infty$: this is Tonelli-Fubini.

If you have a power series
$$f(x):=sum_{k=0}^infty,a_kx^k$$
with radius of convergence $rgeq 1$ ($a_0,a_1,a_2,ldotsinmathbb{C}$), then $f_n|_{[0,1)}to f|_{[0,1)}$ uniformly on compact sets as $ntoinfty$, where $$f_n(x):=sum_{k=0}^n,a_kx^ktext{ for each }xinmathbb{C}text{ and }ninmathbb{Z}_{>0},.$$
This provides a justification for swapping the infinite sum and the integral, that is,
$$int_0^1,f(x),text{d}x=int_0^1,sum_{k=0}^infty,a_kx^k,text{d}x=sum_{k=0}^infty,int_0^1,a_kx^k,text{d}x=sum_{k=0}^infty,frac{a_{k}}{k+1},.$$

In particular, the power series
$$g(x):=sum_{k=0}^infty,chi(k+1),x^k$$
has radius of convergence $dfrac{1}{limsuplimits_{ktoinfty},sqrt[k]{big|chi(k+1)big|}}=1$. Therefore, you can swap the integral and the infinite sum to obtain
$$int_0^1,g(x),text{d}x=int_0^1,sum_{k=0}^infty,chi(k+1),x^k,text{d}x=sum_{k=0}^infty,frac{chi(k+1)}{k+1}=sum_{k=1}^infty,frac{chi(k)}{k},.$$

Note that $x^4,g(x)=g(x)-1+x^2$, so $$g(x)=frac{1-x^2}{1-x^4}=frac{1}{1+x^2}text{ for all }xinmathbb{C}text{ such that }|x|<1,.$$
That is,
$$sum_{k=1}^infty,frac{chi(k)}{k}=int_0^1,frac{1}{1+x^2},text{d}x=arctan(x)big|_{x=0}^{x=1}=frac{pi}{4},.$$

Alternatively, note that
$$chi(k)=frac{text{i}^k-(-text{i})^k}{2text{i}}text{ for each }k=0,1,2,ldots,,$$
where $text{i}$ is the imaginary unit $sqrt{-1}$. From the Taylor series of the principal branch of the natural logarithm function $$ln(1+z)=sum_{k=1}^infty,frac{(-1)^{k-1}}{k},z^k,,$$ we note that the series above converges for $z=pm text{i}$, yielding
$$frac{1}{2},ln(2)+text{i}frac{pi}{4}=ln(1+text{i})=-sum_{k=1}^infty,frac{(-text{i})^k}{k}$$
and
$$frac{1}{2},ln(2)-text{i}frac{pi}{4}=ln(1-text{i})=-sum_{k=1}^infty,frac{text{i}^k}{k},.$$
Subtracting the two equations above and dividing the result by $2text{i}$ yields
$$frac{pi}{4}=sum_{k=1}^infty,frac{text{i}^k-(-text{i})^k}{2text{i},k}=sum_{k=1}^infty,frac{chi(k)}{k},.$$