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the_owner's question at Yahoo! Answers (Convergence of an integral)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello the_owner,

Your integral is $\displaystyle\int_0^{\infty}\dfrac{e^x}{e^{2x}+3}\;dx$. Using the substitution $t=e^x$ we get $$\displaystyle\int_0^{\infty}\dfrac{e^x}{e^{2x}+3}\;dx=\displaystyle\int_1^{\infty}\dfrac{1}{t^2+3}\;dt$$ But $$\displaystyle\lim_{t\to \infty}\left(\dfrac{1}{t^2+1}:\dfrac{1}{t^2}\right)=1\neq 0$$ According to a well-known criterion, the given integral is convergent if and only if $\displaystyle\int_1^{\infty}\dfrac{1}{t^2}\;dt$ is convergent and this one is convergent (again using a well-known theorem), so the given integral is convergent.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
\(\displaystyle \displaystyle \begin{align*} \int_0^{\infty}{\frac{e^x}{e^{2x} + 3}\,dx} = \int_0^{\infty}{\frac{e^x}{ \left( e^x \right) ^2 + 3 }\,dx} \end{align*}\)

Let \(\displaystyle \displaystyle \begin{align*} u = e^x \implies du = e^x\,dx \end{align*}\) and note that \(\displaystyle \displaystyle \begin{align*} u(0) = 1 \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} u \to \infty \end{align*}\) as \(\displaystyle \displaystyle \begin{align*} x \to \infty \end{align*}\), then the integral becomes...

\(\displaystyle \displaystyle \begin{align*} \int_0^{\infty}{\frac{e^x}{\left( e^x \right) ^2 + 3} \, dx} &= \int_1^{\infty}{\frac{1}{u^2 + 3}\,du} \\ &= \lim_{\epsilon \to \infty} \left[ \frac{1}{\sqrt{3}} \arctan{\left( \frac{u}{\sqrt{3}} \right)} \right]_1^{\epsilon} \\ &= \frac{1}{\sqrt{3}}\left\{ \left[ \lim_{ \epsilon \to \infty} \arctan{\left( \frac{\epsilon}{\sqrt{3}} \right) } \right] - \arctan{\left( \frac{1}{\sqrt{3}} \right) } \right\} \\ &= \frac{1}{\sqrt{3}} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) \\ &= \frac{1}{\sqrt{3}} \left( \frac{\pi}{3} \right) \\ &= \frac{\pi \, \sqrt{3}}{9} \end{align*}\)

The integral is obviously convergent...
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
The integral is obviously convergent...
Right, but the problem only asks for the convergence, and this cuestion is more general. For example, to study the convergence of $$\displaystyle\int_1^{\infty}\dfrac{x^3+1}{x^5+11x^4+x^3+\pi x^2+(\log 2)x+\sqrt[3]{2}}\;dx$$ has exactly the same difficulty that to study the convergence of $$\displaystyle\int_1^{\infty}\dfrac{1}{t^2+3} \;dt$$ Another thing, is to compute them.