I'm trying to solve the next integral, using its series. However, I got stuck in a very dumb way nearly at the end.
The infamous: $$\int_{0}^{1} \frac{\text{cosh}(x)-1}{x}dx$$
First, the series of $\text{cosh}(x)$ is: $$\sum_{0}^{\infty}\frac{x^{2n}}{(2n)!}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\text{... etc}$$
Then I replaced the hiperbolic cosine with the series:
$$\int \frac{(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\text{...})-1}{x}dx$$
Reducing and solving the integral:
$$\int(\frac{x}{2!}+\frac{x^3}{4!}+\frac{x^5}{6!}+\text{...})dx=\frac{x^2}{2*2!}+\frac{x^4}{4*4!}+\frac{x^6}{6*6!}+\text{...}+C$$
Then here is where I'm stuck:
$$\frac{x^2}{2*2!}+\frac{x^4}{4*4!}+\frac{x^6}{6*6!}+\text{...}=\sum_{n=1}^{\infty}\frac{x^{2n}}{2n(2n)!}$$
I'm cannot solve the last summation. It seems like a geometric series, but I'm unable to identify its parts, but since the orinigal integral was definite, I need the numerical value.
Is there a way? I feel really dumb when untangling series, and this one seems pretty easy...
TL;DR: $\int_{0}^{1} \frac{\text{cosh}(x)-1}{x}dx =\sum_{n=1}^{\infty}\frac{x^{2n}}{2n(2n)!}|_0^1=?$
For $x=0$, no problem. So, you are concerned by $$S_p=\sum_{n=1}^{p}\frac{1}{2n(2n)!}$$ which will converge very fast as shown below $$\left( \begin{array}{cc} p & S_p \\ 1 & 0.25000000000000000000 \\ 2 & 0.26041666666666666667 \\ 3 & 0.26064814814814814815 \\ 4 & 0.26065124834656084656 \\ 5 & 0.26065127590388007055 \\ 6 & 0.26065127607785304545 \\ 7 & 0.26065127607867238442 \\ 8 & 0.26065127607867537159 \\ 9 & 0.26065127607867538027 \end{array} \right)$$
Edit
Beside the series expansion, you could start using $$I_a=\int_{0}^{a} \frac{\text{cosh}(x)-1}{x}dx=\text{Chi}(a)-\log (a)-\gamma$$ and, instead of using Taylor expansion (which is what you did), use Pade approximants built around $a=0$.
Among the simplest could be $$I_a\approx \frac{\frac{1}{4}a^2+\frac{11 }{4704}a^4}{1-\frac{19 }{588}a^2+\frac{89 }{211680}a^4}\implies I_1\approx\frac{53415}{204929}\approx 0.2606512499 $$