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

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The integral in question, ∫₀^∞ (e^x / (e²ˣ + 3)) dx, is determined to be convergent through a substitution that transforms it into ∫₁^∞ (1 / (t² + 3)) dt. The convergence is established by comparing it to the known convergent integral ∫₁^∞ (1 / t²) dt. The evaluation of the integral leads to a final result of (π√3)/9. The discussion also highlights the complexity of determining convergence for more general integrals, indicating that while the convergence of this specific integral is clear, similar methods apply to other cases.
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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.
 
[math]\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*}[/math]

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

[math]\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*}[/math]

The integral is obviously convergent...
 
Prove It said:
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.
 
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