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an Improper Integral

Jan 31, 2012
54
When the following improper integral converges? When it diverges?


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

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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, \(\displaystyle\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}}\mbox{ should diverge.}\)

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I was trying to solve the integral that seemed unsolvable , actually I didn't read the
question :eek: