How to find Integral $\int_{0}^{1/2} (e^x - 1)/x \,dx$? can you give me any advice how to caluclate $\int_{0}^{1/2} \frac{e^x - 1}{x} \,dx$ ? I need to use series. I tried to split it as
$$
\int_{0}^{1/2} \frac{e^x}{x} \,dx - \int_{0}^{1/2} \frac{1}{x} \,dx$$ and then I can use a series for the first equation but I get another $\int_{0}^{1/2} \frac{1}{x} \,dx$. Can I somehow evaluate these integrals or I need to use a completely different approach. Thanks!
 A: You cannot split the integral into two unless at least one of them is convergent.
By the series definition of the exponential function, for $x\ne 0$
$$
\frac{e^x-1}{x}=\frac1x(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots )
=1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots 
$$
Now define the function $f:[0,1/2]\to\mathbb{R}$ with $f(x)=\frac{e^x-1}{x}$ if $x\ne 0$ and $f(0)=1$. Then the function $f$ is continuous on $[0,1]$.
So you can write
$$
f(x)=1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots
$$
and
$$
\int_0^{1/2} \frac{e^x-1}{x}\,dx=\int_0^{1/2}f(x)dx=
\int_0^{1/2}(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots)\,dx
$$
Now do integration term by term.
A: Hint: $\displaystyle x\ne0\implies\frac{e^x-1}x=\sum_{n=0}^\infty\frac{x^n}{(n+1)!}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1/2}{\expo{x} - 1 \over x}\,\dd x} \,\,\,\stackrel{x\ \mapsto\ -x}{=}\,\,\,
-\int_{-1/2}^{0}{\expo{-x} - 1 \over x}\,\dd x
\\[5mm] = &
\lim_{\Lambda \to \infty}
\bracks{-\int_{-1/2}^{\Lambda}{\expo{-x} - 1 \over x}\,\dd x
+ \int_{0}^{\Lambda}{\expo{-x} - 1 \over x}\,\dd x}
\\[5mm] = &\
\lim_{\Lambda \to \infty}
\left[-{\rm P.V.}\int_{-1/2}^{\Lambda}
{\expo{-x} \over x}\,\dd x\right.
\\[2mm] &\ \phantom{\lim_{\Lambda \to \infty}\,\,\,}
+ \int_{-1/2}^{\Lambda}{\dd x \over x} + 
\ln\pars{\Lambda}\pars{\expo{-\Lambda} - 1}
\\[2mm] &\
\left.\phantom{\lim_{\Lambda \to \infty}\,\,\,}+ \int_{0}^{\Lambda}\ln\pars{x}\expo{-x}\dd x\right]
\\[5mm] = &\
\bbx{\on{E_{i}}\pars{1 \over 2} + \ln\pars{2} - \gamma}
\approx 0.5702 \\ &
\end{align}

*

*$\ds{\on{E_{i}}}$: Exponential Integral Function.
$\ds{\on{E_{i}}\pars{z} \equiv -{\rm P.V.}\int_{-z}^{\infty}{\expo{-x} \over x}\dd x}$.

*$\ds{\gamma}$: Euler-Mascheroni Constant which is equal to $\ds{\phantom{\gamma:}-\int_{0}^{\infty}\ln\pars{x}\expo{-x}\dd x}$.

