Improper integral convergence/divergence

In summary: Now, using the following formula ... $\displaystyle \int_{0}^{\infty} \frac{\cos a x}{1 + x}\ dx = \ln\ \frac{1}{a} + \gamma\ +\ ln\ 2\ (3)$...where $a = 2$ and $\gamma$ is the Euler-Mascheroni constant, You obtain that the second integral of (1) is $- \frac{1}{6}\ (\ln\
  • #1
brunette15
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0
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!
 
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  • #2
brunette15 said:
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.
 
  • #3
brunette15 said:
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?
 
  • #4
brunette15 said:
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 $1$.
 
  • #5
Euge said:
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!
 
  • #6
brunette15 said:
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

$\chi$ $\sigma$
 
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FAQ: Improper integral convergence/divergence

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or when the integrand function is not defined at certain points within the interval of integration.

What are the conditions for convergence of an improper integral?

The two conditions for convergence of an improper integral are that the limit of the integral as the upper or lower limit of integration approaches infinity must exist, and the limit must be a finite value.

What is the comparison test for improper integrals?

The comparison test states that if an improper integral can be shown to be greater than or equal to a convergent integral, then the improper integral must also converge. Similarly, if an improper integral can be shown to be less than or equal to a divergent integral, then the improper integral must also diverge.

What is the limit comparison test for improper integrals?

The limit comparison test is a more powerful version of the comparison test. It states that if the ratio of two improper integrals tends to a finite positive number, then both integrals will either converge or diverge.

How do you determine the convergence or divergence of an improper integral?

To determine the convergence or divergence of an improper integral, you can use one of several tests such as the comparison test, the limit comparison test, the ratio test, or the integral test. If none of these tests apply, you may need to use more advanced methods such as the Cauchy condensation test or the Dirichlet test.

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