How to prove, that integral is divergent $\int_{1}^{\infty} \frac{e^{1/x}-1}{2^{1/x}} \mathrm{d}x$

Question

Let us assume that there is an integral

$$ \int_{1}^{\infty} \frac{e^{\frac{1}{x}}-1}{2^{\frac{1}{x} }} \mathrm{d}x $$

It is necessary to show that this integral diverges

Step 1: Let's make a substitution: $u = \frac{1}{x}$ which gives us: $$ dx = -\frac{du}{u^2} $$

Substituting the variables, the integral becomes:

$$ \int_{0}^{1}\frac{(e^u-1)}{(2^u)} \cdot -\frac{du}{u^2} $$

Splitting the integral into two parts:

\begin{align*} \int_{0}^{1} (e^u) \cdot -\frac{du}{u^2} - \int_{0}^{1} \frac{1}{(2^u)} \cdot -\frac{du}{u^2} \end{align*}

The first integral is diverge. The second integral also is diverge, and if since both integrals diverge, the resulting integral diverges.

Wolfram show, that initial integral is diverge, but I don't understan how to prove this statement. After all, if both integers diverge, this does not mean at all that the final one also diverges. Maybe it is necessary to compare the original sub-integral function with some series?

Answer 1

The integral is improper only because the interval of integration is unbounded, because the integrand function is continuous in $[1,+\infty[$. Hence, noticing that: $$\lim_{x \to +\infty} x\frac{e^{1/x}-1}{2^{1/x}}=1$$ Your integrand function behaves like $1/x$ when $x \to +\infty$, so its integral in $[1,+\infty[$ is divergent.

Another strategy for someone who hasn't studied the limit comparison test yet, but only comparison: notice that $e^t \ge 1+t$ for each $t\in\mathbb{R}$ and that $2^{1/x}$ is decreasing on $[1,+\infty[$, so $\sup_{x \in [1,+\infty[} 2^{1/x}=2$. Hence: $$\int_1^{\infty} \frac{e^{1/x}-1}{2^{1/x}} \text{d}x\ge\int_1^\infty \frac{1+\frac{1}{x}-1}{2}\text{d}x=\frac{1}{2}\int_1^\infty \frac{1}{x}\text{d}x=+\infty$$

Answer 2

To prove that the integral $\int_{1}^{\infty} \frac{e^{1/x}-12}{x} dx$ does not converge, we can use the limit comparison test.

First, let's consider the integral $\int_{1}^{\infty} \frac{1}{x} dx$, which we know diverges. Then, we can compare the given integral to this one by taking the limit as $x$ approaches infinity:

$\lim_{x\to\infty}\frac{x^{-1}e^{1/x}-12}{x^{-1}}=\lim_{x\to\infty}\frac{1}{e^{1/x}}-12=-11$

Since this limit is a finite nonzero number, we can conclude that the given integral behaves the same way as $\int_{1}^{\infty} \frac{1}{x} dx$, which we know diverges. Therefore, the given integral also diverges.

Answer 3

$$ -\int_{0}^{1}\frac{(e^u-1)}{2^u} \cdot \frac{1}{u^2}du $$

Keep going, you are almost done. You want to show this integral diverge, we put the negative sign out of integral, so the integral itself is positive. Next step is to show it approaches to infinity.

Note that: $2^u\le 2,$ for $u\in[0 ,1]$, also we gave $e^u\ge 1+u$, hence

$$\int_{0}^{1}\frac{(e^u-1)}{2^u} \cdot \frac{1}{u^2}du\ge \int_{0}^{1}\frac{u}{2} \cdot \frac{1}{u^2}du=\frac{1}2\ln u~\bigg|_0^1\longrightarrow\infty$$