Does the improper integral $\int_0^{\infty}\frac{2^x+1}{3^x+1}dx$ converge?

Question

Does the following integral converge: $$ \int_0^{\infty}\frac{2^x+1}{3^x+1}dx $$ Using linearity of integrals I did $$ \int_0^{\infty}\frac{2^x+1}{3^x+1}dx = \int_0^{\infty}\frac{2^x}{3^x+1}dx + \int_0^{\infty}\frac{1}{3^x+1}dx $$ The latter can be bounded by improper integral which is known to be convergent. $$ \int_0^{\infty}\frac{1}{3^x+1}dx = \int_0^{\infty}\frac{1}{e^{xln3}+1}dx \sim \int_0^{\infty}\frac{1}{e^{xln3}}dx \le \int_0^{\infty}\frac{1}{e^{x}}dx $$ and as I said $\int_0^{\infty}\frac{1}{e^x}dx$ converges.

What can be done to the first integral?

Answer 1

Your integral converges because$$\lim_{x\to\infty}\frac{\frac{2^x+1}{3^x+1}}{\left(\frac23\right)^x}=1$$and the integral $\int_0^\infty\left(\frac23\right)^x\,\mathrm dx$ converges, since, for each $M>0$,\begin{align}\int_0^M\left(\frac23\right)^x\,\mathrm dx&=\left[\frac{\left(\frac23\right)^x}{\log\left(\frac23\right)}\right]_{x=0}^{x=M}\\&=\frac1{\log\left(\frac23\right)}\left(\left(\frac23\right)^M-1\right)\end{align}and therefore, since $\frac23\in(0,1)$,\begin{align}\int_0^\infty\left(\frac23\right)^x\,\mathrm dx&=\lim_{M\to\infty}\int_0^M\left(\frac23\right)^x\,\mathrm dx\\&=-\frac1{\log\left(\frac23\right)}.\end{align}

Answer 2

Bound it above with $\int_0^\infty\frac{2^xdx}{3^x}$, which converges by the same logic.

Answer 3

Hint:

$\dfrac{2^x}{3^x+1} \lt (2/3)^x.$

Set $e^{-a}= 2/3$, where $a >0$.

Consider $\displaystyle {\int_{0}^{\infty}} e^{-ax}dx .$

Answer 4

Note that \begin{align} \int_0^{\infty}\frac{2^x+1}{3^x+1}dx =& \int_0^{\infty}\frac{2^x}{3^x+1}dx + \int_0^{\infty}\frac{1}{3^x+1}dx \\ \leq& \int_0^{\infty}\frac{2^x}{3^x+1}dx + \int_0^{\infty}\frac{2^x}{3^x+1}dx \\ =& 2\int_0^{\infty}\frac{2^x}{3^x+1}dx \\ \leq& 2\int_0^{\infty}\frac{2^x}{3^x}dx \\ =& 2\cdot \lim_{n\to \infty}\sum_{k=1}^{n} \int_{k-1}^{k}\left(\frac{2}{3}\right)^xdx \\ =& 2\cdot \lim_{n\to \infty}\sum_{k=1}^{n} \int_{k-1}^{k}\left(\frac{2}{3}\right)^{k-1}dx \\ =& 2\cdot \lim_{n\to \infty}\sum_{k=1}^{n} \left(\frac{2}{3}\right)^{k-1} \\ =& 2(3/2)\cdot \lim_{n\to \infty}\sum_{k=1}^{n} \left(\frac{2}{3}\right)^{k} \\ =& 3\cdot \lim_{n\to \infty}\frac{1-(2/3)^{n+1}}{1-2/3} \\ =& 3\cdot\frac{1}{1-2/3} \\ \end{align}