# an Improper Integral

#### Also sprach Zarathustra

##### Member
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
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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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I was trying to solve the integral that seemed unsolvable , actually I didn't read the
question