Study the function: $F(x)=\int_{x}^{2x}\frac{e^{2t}-e^t}{t}\,dt$

Question

Study the function: $$F(x)=\int_{x}^{2x}\frac{e^{2t}-e^t}{t}\,dt$$

I start by saying that it is the first time I am studying a function defined by an Integral. Talking about the domain of this function, we have to discuss the integrability in a neighborhood of $0$ and $\pm \infty$, so I thought to split $F(x)$ into the following sum (if this passage is right):

$$F(x)=\int_{x}^{1}\frac{e^{2t}-e^t}{t}\,dt\,\,+\,\,\int_{1}^{2x}\frac{e^{2t}-e^t}{t}\,dt$$

And then study the convergence of the two. Am I right? Thank you

Answer 1

Some hints:

Note the function $\;\dfrac{\mathrm e^{2t}-\mathrm e^{t}}t$ is Riemann-integrable on every interval $[a,b]$ which doesn't contain $0$. Therefore it is integrable on every interval $[x,2x]$ $(x>0)$ or $[2x,x]$ $(x<0)$, and the domain of the function $F$ is $\mathbf R\smallsetminus\{0\}$.

By the 1st fundamental theorem of integral calculus and the chain rule, its derivative is $$F'(x)=2\,\frac{\mathrm e^{4x}-\mathrm e^{2x}}{2x}-\frac{\mathrm e^{2x}-\mathrm e^{x}}x=\frac{\mathrm e^{4x}-2\mathrm e^{2x}+\mathrm e^{x}}x=\frac{\mathrm e^{x}}x(\mathrm e^{3x}-2\mathrm e^{x}+1),$$ and its sign depends on the sign of $\mathrm e^{3x}-2\mathrm e^{x}+1$. Setting $t=\mathrm e^{x}$, we have to determine the sign of the polynomial $$p(t)=t^3-2t+1$$ on $(0,+\infty)$. This polynomial has $1$ as aroot, so it factors as $$p(t)=(t-1)(t^2+t-1)$$ The quadratic factor has one positive root, $\frac{\sqrt 5-1}2$ and one negative root, $-\frac{\sqrt 5+1}2$. From here, you should be able to determine the monotony of $F$ per intervals.

There would remain to determine the limits of $F(x)$ when $x$ tends to $0$ (use a Taylor expansion at order $1$ near $0$) and when $x$ tends to $\pm\infty$.

Answer 2

You can write this using the Exponential Integral function as $$F(x) = \text{Ei}(x) - 2 \text{Ei}(2x) + \text{Ei}(4x)$$ or, if you prefer, $$ -\text{Ei}_1(-x) + 2 \text{Ei}_1(-2x) - \text{Ei}_1(-4x) $$ Thus for $x$ near $0$ we have a Maclaurin series $$ F(x) = x + \frac{9}{4} x^2 + \frac{49}{18} x^3 + \ldots $$ for $x \to +\infty$ an asymptotic series $$ F(x) \sim \left( \frac{1}{4x} + \frac{1}{16 x^2} + \frac{1}{32 x^3} + \ldots \right) e^{4x} $$ and for $x \to -\infty$ $$ F(x) \sim \left( \frac{1}{x} + \frac{1}{x^2} + \frac{2}{x^3} + \ldots \right) e^{x} $$