Results 1 to 7 of 7
Like Tree4Thanks
  • 1 Post By chiro
  • 1 Post By Prove It
  • 1 Post By Jason76
  • 1 Post By skeeter

Thread: Improper Integral - # 2

  1. #1
    MHF Contributor Jason76's Avatar
    Joined
    Oct 2012
    From
    USA
    Posts
    1,314
    Thanks
    21

    Improper Integral - # 2

    $\displaystyle \int_{0}^{\infty} \dfrac{dx}{x^{2} + 1}$

    $\displaystyle x = (1)\tan\theta$

    $\displaystyle dx = \sec^{2}\theta d\theta$

    $\displaystyle (\tan\theta)^{2} + 1$

    $\displaystyle \tan^{2}\theta + 1 = sec^{2}\theta$

    $\displaystyle \int_{0}^{\infty} \dfrac{\sec^{2}\theta}{\sec^{2}\theta}$ - ??? Simplify further or take integral? Simplification seems to just make a 1 Maybe trig substitution wasn't the way to go.
    Last edited by Jason76; Oct 19th 2014 at 12:51 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    6,608
    Thanks
    1714

    Re: Improper Integral - # 2

    Hey Jason76.

    If you use a trig substitution, you need to remember to change the limits when you do this. Note that your limits should change to [0,pi/2) if you use these kinds of substitutions.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,883
    Thanks
    1948

    Re: Improper Integral - # 2

    Surely you can see that $\displaystyle \begin{align*} \frac{\sec^2{\left( \theta \right)}}{\sec^2{ \left( \theta \right) }} = 1 \end{align*}$...
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Jason76's Avatar
    Joined
    Oct 2012
    From
    USA
    Posts
    1,314
    Thanks
    21

    Re: Improper Integral - # 2

    In this case the answer would be

    $\displaystyle x$ evaluated at $\displaystyle 0$ (lower bound) and $\displaystyle 2 \pi$ upper bound

    $\displaystyle [[x] - [x]]$

    $\displaystyle [2\pi - 0 ]= 2\pi$

    However, I think this problem is another $\displaystyle \arctan(x)$ one as the integral of $\displaystyle \dfrac{1}{x^{2} + 1}$ is $\displaystyle \arctan(x)$

    If your going the $\displaystyle \arctan(x)$ route

    $\displaystyle [\arctan(b)] - [\arctan(0)]$

    $\displaystyle [\arctan(b)] - [0] $

    $\displaystyle \lim b \rightarrow \infty[\arctan(b)] - 0 $ indeterminate

    Using L hopitals

    $\displaystyle \lim b \rightarrow\infty [\dfrac{1}{(\infty)^{2} + 1}]= 0$
    Last edited by Jason76; Oct 19th 2014 at 11:18 AM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Jason76's Avatar
    Joined
    Oct 2012
    From
    USA
    Posts
    1,314
    Thanks
    21

    Re: Improper Integral - # 2

    The answer in the back of the book, though, is $\displaystyle \dfrac{\pi}{2}$

    They must have went the trig substitution route.

    $\displaystyle \int_{0}^{\infty} \dfrac{dx}{x^{2} + 1}$

    $\displaystyle \int_{0}^{\pi/2\dfrac{dx}{x^{2} + 1}$

    $\displaystyle x = (1)\tan\theta$

    $\displaystyle dx = \sec^{2}\theta d\theta$

    $\displaystyle (\tan\theta)^{2} + 1$

    $\displaystyle \tan^{2}\theta + 1 = sec^{2}\theta$

    $\displaystyle \int_{0}^{\infty} \dfrac{\sec^{2}\theta}{\sec^{2}\theta} $

    $\displaystyle \int_{0}^{\pi/2} dx $


    $\displaystyle = x$ evaluated at $\displaystyle 0$ (lower bound) and $\displaystyle \pi/2$ (upper bound)

    $\displaystyle [(\pi/2) - (0)] = \pi/2$ - correct answer on homework.
    Last edited by Jason76; Oct 19th 2014 at 11:26 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3705

    Re: Improper Integral - # 2

    fyi ... $\displaystyle \frac{d}{dx} \arctan{x} = \frac{1}{1+x^2}$ ... you should know this fact.

    so, $\displaystyle \int_0^\infty \frac{dx}{1+x^2} = \lim_{b \to \infty} \left[\arctan{x}\right]_0^b = \lim_{b \to \infty} \left[\arctan(b) - \arctan(0)\right] = \frac{\pi}{2}$

    btw, you are expected to be familiar with the graph of the arctangent function and its end behavior.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Jason76's Avatar
    Joined
    Oct 2012
    From
    USA
    Posts
    1,314
    Thanks
    21

    Re: Improper Integral - # 2

    Quote Originally Posted by skeeter View Post
    fyi ... $\displaystyle \frac{d}{dx} \arctan{x} = \frac{1}{1+x^2}$ ... you should know this fact.

    so, $\displaystyle \int_0^\infty \frac{dx}{1+x^2} = \lim_{b \to \infty} \left[\arctan{x}\right]_0^b = \lim_{b \to \infty} \left[\arctan(b) - \arctan(0)\right] = \frac{\pi}{2}$

    btw, you are expected to be familiar with the graph of the arctangent function and its end behavior.
    $\displaystyle \dfrac{\pi}{2}$ would have came as the answer using either method.

    The limit of $\displaystyle \arctan(x)$ as it approaches infinity is $\displaystyle \pi/2$
    Last edited by Jason76; Oct 19th 2014 at 11:36 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Improper Integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jun 3rd 2013, 06:28 PM
  2. Improper Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 8th 2012, 08:38 PM
  3. Improper Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 11th 2011, 10:03 AM
  4. [SOLVED] Improper integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 24th 2011, 01:38 AM
  5. Improper integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 10th 2009, 02:41 AM

Search Tags


/mathhelpforum @mathhelpforum