MHB Improper Integral Convergence & Divergence

AI Thread Summary
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
Messages
43
Reaction score
0
When the following improper integral converges? When it diverges? $$ \int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)} $$
 
Mathematics news on Phys.org
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, \(\displaystyle\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}}\mbox{ should diverge.}\)

\[\therefore\displaystyle\int^{\infty}_{0} \frac{x^{\alpha}dx}{1+x^{\beta}\sin^2(x)}\mbox{ diverges when }\beta>0\mbox{ and }\alpha+1>0\]
 
Last edited:
I was trying to solve the integral that seemed unsolvable , actually I didn't read the
question :o
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top