Evaluate $\int_{a}^{b} \frac{\left(e^{\frac{x}{a}}-e^{\frac{b}{x}}\right)dx}{x}$

Question

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?

Answer 1

Hint: Let $x=\dfrac{ab}{u}$ then $$g(b)=\int_{a}^{b} \frac{e^{\frac{b}{x}}dx}{x}=f(a)$$

Answer 2

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}

Answer 3

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.