Results 1 to 4 of 4

Math Help - Improper Integral Help

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    7

    Improper Integral Help

    Just needed a bit of confirmation that I was doing this problem correctly. I started off using a table definition that you can see at the bottom of that picture. The part that got cut off just shows: u/a^2sqrt(u^2+a^2). After that I just put in 0 and infinity, and came out with the constants * 1/r^2. Does that seem like an appropriate answer?

    Thanks!

    Link: imgur: the simple image sharer
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294

    Re: Improper Integral Help

    Quote Originally Posted by masteroc View Post
    Just needed a bit of confirmation that I was doing this problem correctly. I started off using a table definition that you can see at the bottom of that picture. The part that got cut off just shows: u/a^2sqrt(u^2+a^2). After that I just put in 0 and infinity, and came out with the constants * 1/r^2. Does that seem like an appropriate answer?

    Thanks!

    Link: imgur: the simple image sharer
    First you need to find the indefinite integral, then see if it converges.

    Let \displaystyle x = r\tan{\theta} \implies dx = r\sec^2{\theta}\,d\theta and the integral becomes

    \displaystyle \begin{align*}\int{\frac{dx}{(r^2 + x^2)^{\frac{3}{2}}}} &= \int{\frac{r\sec^2{\theta}\,d\theta}{[r^2 + (r\tan{\theta})^2]^{\frac{3}{2}}}} \\ &= \int{\frac{r\sec^2{\theta}\,d\theta}{[r^2 + r^2\tan^2{\theta}]^{\frac{3}{2}}}} \\ &= \frac{r\sec^2{\theta}\,d\theta}{[r^2(1 + \tan^2{\theta})]^{\frac{3}{2}}} \\ &= \frac{r\sec^2{\theta}\,d\theta}{(r^2\sec^2{\theta}  )^{\frac{3}{2}}} \\ &= \frac{r\sec^2{\theta}\,d\theta}{r^3\sec^3{\theta}} \\ &= \int{\frac{d\theta}{r^2\sec{\theta}}} \\ &= \frac{1}{r^2}\int{\cos{\theta}\,d\theta} \\ &= \frac{\sin{\theta}}{r^2} + C \\ &= \frac{\tan{\theta}}{r^2\sqrt{1 + \tan^2{\theta}}} + C \\ &= \frac{\frac{x}{r}}{r^2\sqrt{1 + \frac{x^2}{r^2}}} + C \\ &= \frac{x}{r^3\sqrt{\frac{r^2 + x^2}{r^2}}} + C \\ &= \frac{x}{r^2\sqrt{r^2 + x^2}} + C\end{align*}

    So now to take the definite integral between \displaystyle 0 and \displaystyle \infty, we need to evaluate

    \displaystyle \lim_{\epsilon \to \infty}\left(\frac{\epsilon}{r^2\sqrt{r^2 + \epsilon^2}}\right) - \frac{0}{r^2\sqrt{r^2 + 0^2}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    7

    Re: Improper Integral Help

    Thanks for the reassurance, I got that answer using the tables definition and when I went to evaluate, I got the right part of that (the non-limit part) equaling zero. However, the limit part i'm confused with. The top part is infinity and so is the bottom it would seem. I used LHopitals Rule, but it seemed to keep on going forever because the bottom always equaled infinity which would make this problem 0-0=0....and that does not seem correct given that this is a physics formula and usually would have an answer greater than 0.

    Thanks again for the quick response!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294

    Re: Improper Integral Help

    Quote Originally Posted by masteroc View Post
    Thanks for the reassurance, I got that answer using the tables definition and when I went to evaluate, I got the right part of that (the non-limit part) equaling zero. However, the limit part i'm confused with. The top part is infinity and so is the bottom it would seem. I used LHopitals Rule, but it seemed to keep on going forever because the bottom always equaled infinity which would make this problem 0-0=0....and that does not seem correct given that this is a physics formula and usually would have an answer greater than 0.

    Thanks again for the quick response!
    L'Hospital's Rule isn't needed...

    \displaystyle \begin{align*} \frac{\epsilon}{r^2\sqrt{r^2 + \epsilon^2}} &= \frac{\epsilon \cdot \frac{1}{\epsilon}}{r^2\sqrt{r^2 + \epsilon ^2}\cdot \frac{1}{\epsilon}} \\ &= \frac{1}{\frac{r^2\sqrt{r^2 +\epsilon ^2}}{\sqrt{\epsilon^2}}} \\ &= \frac{1}{r^2\sqrt{\frac{r^2 + \epsilon ^2}{\epsilon ^2}}} \\ &= \frac{1}{r^2\sqrt{\frac{r^2}{\epsilon^2} + 1}} \end{align*}

    Now make \displaystyle \epsilon \to \infty...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Improper integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 19th 2010, 01:19 PM
  2. Improper Integral help?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 15th 2010, 04:03 PM
  3. Improper Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 27th 2010, 11:58 AM
  4. Improper integral 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 26th 2008, 10:54 AM
  5. Improper integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 25th 2008, 11:14 PM

Search Tags


/mathhelpforum @mathhelpforum