$\int_1^\infty 1/\sqrt{1+e^x}dx$

Question

Using the comparison test I am supposed to figure out whether this integral converges or diverges. what other function should I use? Also, the inequality stating that $1/\sqrt{e^x+1}$ is larger or smaller than another function must be proven.

Answer 1

First, why is there any question of convergence or divergence? Check that the integrand is continuous and defined for all real numbers, so there's no vertical asymptotes to be concerned about; that leaves only the upper bound of $\infty$.

Next, let's guess whether it converges or not. For this, we try to understand what happens to the integrand as $x\to\infty$. Then $e^x\to\infty$, but 1 is just 1 ($e^x$ is "big", and 1 is "regular size"). So $1+e^x\approx e^x$, and $$ \frac1{\sqrt{1+e^x}}\approx\frac1{\sqrt{e^x}} = e^{-x/2} $$ That's an integrand you could just integrate directly; so do that, and verify that $\int_1^\infty e^{-x/2}\,dx$ is finite.

At this point we suspect that the original integral converges. To prove that, we'll use the comparison theorem, as you said. The integrand is positive, and we expect to show that it converges, so we're looking for an upper bound: $$ \frac1{\sqrt{1+e^x}} \le g(x) $$ where $\int_1^\infty g(x)\,dx$ is finite. Now, if we're lucky, the same approximation we did above will work as an inequality in the direction we need. It turns out that in this case, we're lucky: since $1\ge 0$, we have $\sqrt{1+e^x}\ge\sqrt{0+e^x}=e^{x/2}$, and so $$ \frac1{\sqrt{1+e^x}} \le e^{-x/2} $$ and off we go.

(If it had turned out that we wanted the opposite inequality, which will happen in other problems of this type, then you could get one as follows: for $x\ge0$ we have $1\le e^x$, so $\sqrt{1+e^x}\le\sqrt{e^x+e^x} = \sqrt2 e^{x/2}$, and $$ \frac1{\sqrt{1+e^x}} \ge \frac1{\sqrt2} \,e^{-x/2} $$ This would be good enough for the purposes of the comparison test, since the $\frac1{\sqrt2}$ will just pop out in front of the integral and not affect whether it comes out to be finite or infinite. This is a common trick: when we were just approximating, we threw away the 1 as negligibly small in context; but if doing that yields an inequality in the wrong direction, then rather than throw it away, replace it with a copy of the nearby large terms and combine them, perhaps at the cost of a constant factor.)

Answer 2

Hint: Clearly, $1+e^x \ge e^x$, so $\sqrt{1+e^x} \ge \sqrt{e^x} = e^{x/2}$, and thus, $\dfrac{1}{\sqrt{1+e^{x}}} \le \dfrac{1}{e^{x/2}} = e^{-x/2}$.

Using the comparison test should be easy now.

Answer 3

Note that we can calculate the precise value of the integral, not only the convergence. We have $$\int_{1}^{\infty}\frac{1}{\sqrt{1+e^{x}}}dx\overset{{\scriptstyle e^{x}=u}}{=}\int_{e}^{\infty}\frac{1}{\sqrt{u+1}u}du\overset{{\scriptstyle u+1=v}}{=}\int_{1+e}^{\infty}\frac{1}{\left(v-1\right)\sqrt{v}}dv\overset{{\scriptstyle t=\sqrt{v}}}{=}2\int_{\sqrt{1+e}}^{\infty}\frac{1}{t^{2}-1}dt $$ and now the integral is easy to solve $$=\lim_{a\rightarrow\infty}\left.\left(-2\tanh^{-1}\left(t\right)\right)\right|_{\sqrt{1+e}}^{a}=2\coth^{-1}\left(\sqrt{1+e}\right). $$