Evaluate $\int_{a}^{b} \frac{\left(e^{\frac{x}{a}}-e^{\frac{b}{x}}\right)dx}{x}$ Evaluate $$I(a,b)=\int_{a}^{b} \frac{\left(e^{\frac{x}{a}}-e^{\frac{b}{x}}\right)dx}{x}$$ given $a,b \in \mathbb{R^+}$
My attempt:
we have $$I(a,b)=f(a)-g(b)$$ where 
$$f(a)=\int_{a}^{b} \frac{e^{\frac{x}{a}}dx}{x}$$ and
$$g(b)=\int_{a}^{b} \frac{e^{\frac{b}{x}}dx}{x}$$
differentiating $f(a)$with respect to $a$ we get
$$f'(a)=\int_{a}^{b}e^{\frac{x}{a}} \times \frac{-x dx}{a^2x}=\frac{e-e^{\frac{b}{a}}}{a} \tag{1}$$
Differentiating $g(b)$with respect to $b$ we get
$$g'(b)=\frac{e^{\frac{b}{a}}-e}{b} \tag{2}$$
Hence
$$af'(a)+bg'(b)=0$$
Any way to proceed here?
 A: Hint:
Let $x=\dfrac{ab}{u}$ then
$$g(b)=\int_{a}^{b} \frac{e^{\frac{b}{x}}dx}{x}=f(a)$$
A: Let $x=\sqrt{ab}\,e^z$. Then
$$ I(a,b) = \int_{a}^{b}\left(e^{x/a}-e^{b/x}\right)\frac{dx}{x} = \sqrt{ab}\int_{-\frac{1}{2}\log\frac{b}{a}}^{\frac{1}{2}\log\frac{b}{a}}\exp\left(e^z\sqrt{b/a}\right)-\exp\left(e^{-z}\sqrt{b/a}\right)\,dz $$
clearly equals zero since it is the integral of an odd integrable function over a symmetric interval with respect to the origin.
A: Let's solve the integral of the form
\begin{equation}
 \int \frac{e^{x/a}}{x}
\end{equation}
Use the change of variable
\begin{equation}
 u = \frac{x}{a}
\end{equation}
you get
\begin{equation}
 dx = a du
\end{equation}
so the integral becomes
\begin{equation}
 \int \frac{e^{u}}{u}
\end{equation}
The above is known as the exponential integral i.e.
\begin{equation}
 \int \frac{e^{u}}{u}
 =
 \text{Ei}(u) + K
\end{equation}
If you would like to know more about this function, let's use Taylor series to expand $e^{u}$
\begin{equation}
 e^u = \sum\limits_{n=0}^{\infty} \frac{u^n}{n!}
\end{equation}
So,
\begin{equation}
 \frac{e^u}{u} = 
 \frac{1}{u}
 +
\sum\limits_{n=0}^{\infty} \frac{u^n}{(n+1)!}
\end{equation}
Hence
\begin{equation}
 \int \frac{e^{u}}{u}
 =
 \int \frac{1}{u}
 +
\sum\limits_{n=0}^{\infty} \int \frac{u^n}{(n+1)!}
=
\ln( \vert u \vert ) 
+
\sum\limits_{n=1}^{\infty}
\frac{u^n}{n . n!}
+
K
\end{equation}
