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

[Fernando Revilla]

Here is the question:

I uploaded it here:
http://i47.tinypic.com/m8moo4.png

Thanks!

Here is a link to the question:

Determine if the integral is convergent or divergent? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

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]

\(\displaystyle \displaystyle \begin{align*} \int_0^{\infty}{\frac{e^x}{e^{2x} + 3}\,dx} = \int_0^{\infty}{\frac{e^x}{ \left( e^x \right) ^2 + 3 }\,dx} \end{align*}\)

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]

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.

 

[CellCoree]

Hello,

Thank you for providing the link to your integral question. After reviewing the image, I can confirm that the integral is convergent. This can be determined by using the limit comparison test or the direct comparison test, both of which compare the given integral to a known convergent or divergent integral.

If you need further clarification or assistance with solving the integral, please feel free to ask follow-up questions. Thank you for utilizing online resources for your mathematical inquiries.

Best,

Scientist