MHB Improper Integral Convergence & Divergence

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The convergence of the improper integral $$ \int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)} $$ depends on the values of α and β. It diverges when β < 0 for any α. When β > 0 and α + 1 < 0, the integral converges, while it diverges if β > 0 and α + 1 > 0. The Limit Comparison Test is used to analyze the behavior of the integral in these cases. Understanding these conditions is crucial for determining the integral's convergence or divergence.
Also sprach Zarathustra
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When the following improper integral converges? When it diverges? $$ \int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)} $$
 
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Also sprach Zarathustra said:
When the following improper integral converges? When it diverges?

$$ \int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)} $$

Hi Also sprach Zarathustra,

\[\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\]

We shall use the Limit Comparison Test to determine the convergence/divergence of this improper integral.

Let, \(\displaystyle f(x)=\frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)} \mbox{ and }g(x)=x^{\alpha}\). It is clear that both \(f(x)\mbox{ and }g(x)\) are positive for all \(x>0\).

Case 1: When \(\mathbf{\beta<0}\)

\[\displaystyle\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=\lim_{x\rightarrow\infty}\frac{1}{1+x^{\beta}\sin^2(x)}=1\]

It is clear that, \(\displaystyle\int_{0}^{\infty}x^{\alpha}\,dx\) diverges for each \(\alpha\in\Re\)

\[\therefore\displaystyle\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\mbox{ diverges when }\beta<0\]

Case 2: When \(\mathbf{\beta>0\mbox{ and }\alpha+1<0}\)

\[\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}=\int^{1}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}+\int^{ \infty}_{1} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\]

Since \(\displaystyle\int^{1}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\) is a proper integral, converge/divergence of \(\displaystyle\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\) depends on the convergence/divergence of \(\displaystyle\int^{\infty}_{1} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\)

\[\displaystyle\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=\lim_{x\rightarrow\infty}\frac{1}{1+x^{\beta}\sin^2(x)}=0\]

\(\displaystyle\int_{1}^{\infty}x^{\alpha}\,dx=-\frac{1}{\alpha+1}\mbox{ for each }\alpha+1<0\)

\[\therefore\int^{\infty}_{1}\frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\mbox{ converges when }\beta>0\mbox{ and }\alpha+1<0\]

\[\Rightarrow\int^{\infty}_{0}\frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\mbox{ converges when }\beta>0\mbox{ and }\alpha+1<0\]

Case 3: When \(\mathbf{\beta>0\mbox{ and }\alpha+1>0}\)

For this case I need a little bit of help from the Wolfram Integrator. :)

It could be shown that, \(\displaystyle \frac{x^{ \alpha}}{1+x^{\beta}\sin^2(x)}>\frac{x^{ \alpha}}{1+x^{\beta}}\mbox{ for }x>0\,.\)

For \(\displaystyle\int\frac{x^{\alpha}dx}{1+x^{\beta}}\) the Wolfram Integrator gives,

\[\displaystyle\int\frac{x^{\alpha}dx}{1+x^{\beta}}=\frac{x^{ \alpha+1}\,_2F_1\left(1,\frac{ \alpha+1}{ \beta},\frac{\alpha+1}{ \beta}+1,-x^{\beta}\right)}{\alpha+1}\]

Where \(\,_2F_1\) is the Hypergeometric series.

\[\Rightarrow\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}}=\lim_{x \rightarrow\infty}\left\{\frac{x^{ \alpha+1}\,_2F_1\left(1,\frac{ \alpha+1}{ \beta},\frac{\alpha+1}{ \beta}+1,-x^{\beta}\right)}{\alpha+1}\right\}-\left\{\frac{x^{ \alpha+1}\,_2F_1\left(1,\frac{ \alpha+1}{ \beta},\frac{\alpha+1}{ \beta}+1,0\right)}{\alpha+1}\right\}\]

Since the \(x^{ \alpha+1}\) term explodes as \(x\rightarrow\infty\) the first term in the right hand side diverges. The radius of convergence of the Hypergeometric series is 1 and therefore the second term has a finite value. Hence, \(\displaystyle\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}}\mbox{ should diverge.}\)

\[\therefore\displaystyle\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\mbox{ diverges when }\beta>0\mbox{ and }\alpha+1>0\]
 
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I was trying to solve the integral that seemed unsolvable , actually I didn't read the
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