Determine some properties of the integral of et / t

Let

$F(x) = \int_1^x \frac{e^t}{t} \, dt \qquad \text{for} \quad x > 0.$

Find all values of $x > 0$ such that $log x \leq F(x)$ .
Prove that
$\int_1^x \frac{e^t}{t+a} \, dt = e^{-a} (F(x+a) - F(1+a)).$
Similar to part (b), find expressions for
$\int_1^x \frac{e^{at}}{t} \, dt, \qquad \int_1^x \frac{e^t}{t^2} \, dt, \qquad \int_1^x \e^{\frac{1}{t}} \, dt.$

Recalling the definition of $\log x$ as the integral,
$\log x = \int_1^x \frac{1}{t} \, dt,$

We compute,

$F(x) - \log x = \int_1^x \frac{e^t}{t} \, dt - \int_1^x \frac{1}{t} \, dt = \int_1^x \frac{e^t -1}{t} \, dt.$

Then, since

$\frac{e^t - 1}{t} \geq 0 \qquad \text{for all } t > 0,$

we have

$\int_1^x \frac{e^t - 1}{t} \, dt \geq 0 \qquad \text{if} \quad x \geq 1,$

and

$\int_1^x \frac{e^t - 1}{t} \, dt = - \int_x^1 \frac{e^t - 1}{t} \, dt < 0 \qquad \text{if} \quad x < 1.$

Hence, the set of $x$ such that $\log x \leq F(x)$ is all real $x \geq 1$ .
Proof. We start by computing,
$\begin{align*} \int_1^x \frac{e^t}{t+a} &= \frac{1}{e^a} \int_1^x \frac{e^t e^a}{t+a} \, dt \\[9pt] &= e^{-a} \int_1^x \frac{e^{t+a}}{t+a} \, dt. \end{align*}$

Then, using the translation invariance of the integral,

$\begin{align*} e^{-a} \int_1^x \frac{e^{t+a}}{t+a} \, dt &= e^{-a} \int_{1+a}^{x+a} \frac{e^{(t-a)+a}}{(t-a)+a} \, dt \\[9pt] &= e^{-a} \int_{1+a}^{x+a} \frac{e^t}{t} \, dt \\[9pt] &= e^{-a} \left( \int_1^{x+a} \frac{e^t}{t} \, dt - \int_1^{1+a} \frac{e^t}{t} \, dt \right) \\[9pt] &= e^{-a} (F(x+a) - F(1+a)). \qquad \blacksquare \end{align*}$
We claim the following formulas hold:
$\begin{align*} \int_1^x \frac{e^{at}}{t} \, dt &= F(ax) - F(a) \\[10pt] \int_1^x \frac{e^t}{t^2} \, dt &= F(x) - \frac{e^x}{x} + e \\[10pt] \int_1^x e^{\frac{1}{t}} \, dt &= xe^{frac{1}{x}} - e - F \left( \frac{1}{x} \right). \end{align*}$

Proof. Making the substitution $s = at$ , $ds = a \, dt$ we have

$\begin{align*} \int_1^x \frac{e^{at}}{t} \, dt &= \int_a^{ax} \frac{e^s}{\frac{s}{a}} \frac{1}{a} \, ds \\[9pt] &= \int_a^{ax} \frac{e^s}{s} \, ds \\[9pt] &= \int_1^{ax} \frac{e^s}{s} \, ds - \int_1^a \frac{e^s}{s} \, ds \\[9pt] &= F(ax) - F(a). \qquad \blacksquare \end{align*}$

Proof. To prove this identity we integrate by parts, letting

$\begin{align*} u &= e^t & \implies && du &= e^t \, dt \\ dv &= \frac{1}{t^2} \, dt & \implies && v &= -\frac{1}{t}. \end{align*}$

Therefore,

$\begin{align*} \int_1^x \frac{e^t}{t^2} \, dt &= \left. -\frac{e^t}{t}\right|_1^x + \int_1^x \frac{e^t}{t} \, dt \\[9pt] &= e - \frac{e^x}{x} + F(x) \\[9pt] &= F(x) - \frac{e^x}{x} + e. \qquad \blacksquare \end{align*}$

Proof. For the final identity, we use the substitution $s = \frac{1}{t}$ , $ds = -\frac{1}{t^2} \, dt$ . This gives us

$\int_1^x e^{\frac{1}{t}} \, dt &= -\int_1^{\frac{1}{x}} \frac{e^s}{s^2} \, ds.$

Now, we use integration by parts with

$\begin{align*} u &= e^s & \implies && du &= e^s \, ds \\ dv &= \frac{1}{s^2} \, ds & \implies && v &= -\frac{1}{s}. \end{align*}$

This gives us

$\begin{align*} \int_1^x e^{\frac{1}{t}} \, dt &= - \int_1^{\frac{1}{x}} \frac{e^s}{s^2} \, ds \\[9pt] &= \frac{e^s}{s} \Bigr \rvert_1^{\frac{1}{x}} - \int_1^{\frac{1}{x}} \frac{e^s}{s} \, ds \\[9pt] &= xe^{\frac{1}{x}} - e - F \left( \frac{1}{x} \right). \qquad \blacksquare \end{align*}$