MHB Improper integral convergence/divergence

[brunette15]

I am attempting to solve the improper integral (x*cos^2(x))/(1+x^3) dx between infinity and 1 to see if it converges or diverges. My approach was to place a point 'x' that approaches infinity to be able to solve the integral and then evaluate the limits however i am stuck on actually computing the antiderivative. Can anyone help me with this?
Thanks in advance!

 

[Opalg]

I am attempting to solve the improper integral (x*cos^2(x))/(1+x^3) dx between infinity and 1 to see if it converges or diverges. My approach was to place a point 'x' that approaches infinity to be able to solve the integral and then evaluate the limits however i am stuck on actually computing the antiderivative. Can anyone help me with this?
Thanks in advance!

You will not be able to find this antiderivative explicitly. What you need to do is to use a comparison test. If the function $(x\cos^2(x))/(1+x^3)$ is smaller than some function whose integral converges, then this integral will also converge. On the other hand, if the function $(x\cos^2(x))/(1+x^3)$ is larger than some function whose integral diverges, then this integral will also diverge.

 

[Prove It]

I am attempting to solve the improper integral (x*cos^2(x))/(1+x^3) dx between infinity and 1 to see if it converges or diverges. My approach was to place a point 'x' that approaches infinity to be able to solve the integral and then evaluate the limits however i am stuck on actually computing the antiderivative. Can anyone help me with this?
Thanks in advance!

Notice that $\displaystyle \begin{align*} \frac{x\cos^2{(x)}}{1 + x^3} \geq 0 \end{align*}$ for all $\displaystyle \begin{align*} x \geq 1 \end{align*}$, and since $\displaystyle \begin{align*} 0 \leq \cos^{(x)} \leq 1 \end{align*}$ for all $\displaystyle \begin{align*} x \end{align*}$, that means $\displaystyle \begin{align*} \frac{x \cos^2{(x)}}{1 + x^3} \leq \frac{x}{1 + x^3} \end{align*}$ for all $\displaystyle \begin{align*} x \geq 1 \end{align*}$. Thus we can say for sure $\displaystyle \begin{align*} 0 \leq \int_1^{\infty}{\frac{x\cos^2{(x}}{1 + x^3}\,\mathrm{d}x} \leq \int_1^{\infty}{\frac{x}{1 + x^3}\,\mathrm{d}x} \end{align*}$. Thus, if $\displaystyle \begin{align*} \int_1^{\infty}{\frac{x}{1 + x^3}\,\mathrm{d}x} \end{align*}$ is convergent, so is $\displaystyle \begin{align*} \int_1^{\infty}{\frac{x\cos^2{(x)}}{1 +x^3}\,\mathrm{d}x} \end{align*}$. So looking at the new integral, we have

$\displaystyle \begin{align*} \int_1^{\infty}{\frac{x}{1 + x^3}\,\mathrm{d}x} &= \int_1^{\infty}{ \frac{x}{\left( 1 + x \right) \left( 1 - x + x^2 \right) } \,\mathrm{d}x } \end{align*}$

Now applying partial fractions, we have

$\displaystyle \begin{align*} \frac{A}{1 + x} + \frac{B\,x + C}{1 - x + x^2} &\equiv \frac{x}{ \left( 1 + x \right) \left( 1 - x + x^2 \right) } \\ \frac{A \left( 1 - x + x^2 \right) + \left( B\,x + C \right) \left( 1 + x \right) }{\left( 1 + x \right) \left( 1 - x + x^2 \right) } &\equiv \frac{x}{ \left( 1 + x \right) \left( 1 - x + x^2 \right) } \\ A \left( 1 - x + x^2 \right) + \left( B\,x + C \right) \left( 1 + x \right) &\equiv x \end{align*}$

Now let $\displaystyle \begin{align*} x = -1 \end{align*}$ and we find $\displaystyle \begin{align*} 3A = -1 \implies A = -\frac{1}{3} \end{align*}$. Substituting in gives

$\displaystyle \begin{align*} -\frac{1}{3} \left( 1 - x + x^2 \right) + \left( B\,x + C \right) \left( 1 + x \right) &\equiv x \\ -\frac{1}{3} + \frac{1}{3}x - \frac{1}{3}x^2 + B\,x + B\,x^2 + C + C\,x &\equiv x \\ \left( B - \frac{1}{3} \right) \, x^2 + \left( B + C + \frac{1}{3} \right) \, x + C - \frac{1}{3} &\equiv 0x^2 + 1x + 0 \end{align*}$

Equating coefficients we find $\displaystyle \begin{align*} C - \frac{1}{3} = 0 \implies C = \frac{1}{3} \end{align*}$ and $\displaystyle \begin{align*} B - \frac{1}{3} = 0 \implies B = \frac{1}{3} \end{align*}$. Thus

$\displaystyle \begin{align*} \int_1^{\infty}{\frac{x}{\left( 1 + x \right) \left( 1 - x + x^2 \right) }\,\mathrm{d}x} &= \int_1^{\infty}{ \frac{-\frac{1}{3}}{1 + x} + \frac{\frac{1}{3} x + \frac{1}{3}}{1 -x + x^2} \,\mathrm{d}x } \\ &= \frac{1}{3} \int_1^{\infty}{ \frac{1}{x^2 - x + 1} - \frac{1}{x +1} \, \mathrm{d}x } \\ &= \frac{1}{3} \int_1^{\infty}{ \frac{1}{ \left( x - \frac{1}{2} \right) ^2 + \frac{3}{4} } - \frac{1}{x + 1} \,\mathrm{d}x } \end{align*}$

Can you continue?

 

[Euge]

I am attempting to solve the improper integral (x*cos^2(x))/(1+x^3) dx between infinity and 1 to see if it converges or diverges. My approach was to place a point 'x' that approaches infinity to be able to solve the integral and then evaluate the limits however i am stuck on actually computing the antiderivative. Can anyone help me with this?
Thanks in advance!

Hi brunette15,

Note that

$$0 \le \frac{x\cos^2 x}{1 + x^3} \le \frac{1}{x^2}\quad \text{for all} \quad x \ge 1.\tag{*}$$

To get the last inequality, use the upper bound $\cos^2 x \le 1$ and the inequality $1 + x^3 \ge x^3$. Now since $(*)$ holds, it suffices (by the comparison test for improper integrals) to show that $\int_1^\infty \frac{dx}{x^2}$ converges. Compute $\int_1^\infty \frac{dx}{x^2}$ directly to show that it converges to $1$.

 

[brunette15]

Hi brunette15,

Note that

$$0 \le \frac{x\cos^2 x}{1 + x^3} \le \frac{1}{x^2}\quad \text{for all} \quad x \ge 1.\tag{*}$$

To get the last inequality, use the upper bound $\cos^2 x \le 1$ and the inequality $1 + x^3 \ge x^3$. Now since $(*)$ holds, it suffices (by the comparison test for improper integrals) to show that $\int_1^\infty \frac{dx}{x^2}$ converges. Compute $\int_1^\infty \frac{dx}{x^2}$ directly to show that it converges to $1$.

Thankyou everyone!

 

[chisigma]

I am attempting to solve the improper integral (x*cos^2(x))/(1+x^3) dx between infinity and 1 to see if it converges or diverges. My approach was to place a point 'x' that approaches infinity to be able to solve the integral and then evaluate the limits however i am stuck on actually computing the antiderivative. Can anyone help me with this?
Thanks in advance!

Let suppose that the integral is from 0 to $\infty$, so that we have to compute...

$\displaystyle I = \int_{0}^{\infty} \frac{x\ \cos^{2} x}{1+x^{3}}\ dx = \frac{1}{2}\ \int_{0}^{\infty} \frac{x}{1+x^{3}}\ d x + \frac{1}{2}\ \int_{0}^{\infty} \frac{x\ \cos 2 x}{1+x^{3}}\ d x\ (1)$

The first integral is a 'not impossible task' and You obtain...

$\displaystyle \frac{1}{2}\ \int_{0}^{\infty} \frac{x}{1+x^{3}}\ dx = \frac{\pi}{3\ \sqrt{3}}\ (2)$

The second integral is a little more complex ...

$\displaystyle \frac{1}{2}\ \int_{0}^{\infty} \frac{x\ \cos 2x}{1+x^{3}}\ dx = - \frac{1}{6}\ \int_{0}^{\infty} \frac{\cos\ 2 x}{1 + x}\ dx + \frac{1}{12}\ \int_{0}^{\infty} \frac{(1 + i\ \sqrt{3})\ \cos 2 x}{x - \frac{1 + i\ \sqrt{3}}{2}}\ d x +\frac{1}{12}\ \int_{0}^{\infty} \frac{(1 - i\ \sqrt{3})\ \cos 2 x}{x - \frac{1 - i\ \sqrt{3}}{2}}\ d x\ (3) $

And now?... now You can use the relation found in...

http://mathhelpboards.com/analysis-50/too-difficult-integral-1842.html?highlight=difficult+integral

$\displaystyle \int_{0}^{\infty} \frac{\cos \omega x}{x + a}\ d x = - \cos (a\ \omega)\ \text{Ci}\ (a\ \omega) + \sin (a\ \omega)\ [\frac{\pi}{2} - \text{Si}\ (a\ \omega)]\ (4) $

Of course the effective computation is not very comfortable, expecially for me, very poor in pure calculus (Wasntme) ...

Kind regards

$\chi$ $\sigma$