Integral of inverse tangent

Discussion in 'Math' started by Razor Concepts, Dec 4, 2009.

  1. Razor Concepts

    Thread Starter Active Member

    Oct 7, 2008
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    I am in AB calculus, however I need to find the intergral of an inverse tangent function for a project. I have attached what I had gotten so far. I retrieved the integral of the inverse tangent of x from yahoo answers, the person who solved it used some kind of integrating by parts method.

    But the function I really need to get the integral of is also attached below. Maybe someone can pop it in a TI-89 and get the integral for me? Also I am wondering what level of math this kind of thing is? Thanks!

    atan is inverse tangent.
    R and r are just other variables

    [​IMG]
    http://i46.tinypic.com/md1eup.jpg
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    You need to think a bit harder about functions when trying to do this integration.
    Sketch the graph of y = atanx

    Put x = 0 into your first formula, what do you get?

    The function goes to infinity on any interval which includes the origin.

    So the first question is: what are the limits of integration?

    In your second formula you say R and r are other variables.
    How can anyone integrate them on this information?
    What are they they in terms of the variable of integration?
     
  3. steveb

    Senior Member

    Jul 3, 2008
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    I agree with studiot about the importance of clearly defining the question.

    Two things bother me in the notation shown. First, the letter "a" should be in normal font to match the font used for "tan". If you don't do this, then someone will think you mean a constant "a" time the normal tangent function.
    The standard is to use italics for variables and normal font for functions. Second, you need to show the dx so that it is clear that the variable of integration is "x".

    Then studiot's point is important. The symbols "r" and "R" must either be representing constants, or variables that are independent of the variable "x". Otherwise, the functional dependence of R(x) and r(x) must be specified in order to solve the problem.

    If we assume that "r" and "R" are constants and that the variable of integration is in fact "x", then the integral can be expressed in closed form notation by using a variable substitution, and integral table, as follows.

    Let  u={{R}\over{r\; x}}

    Next, find dx in terms of du and u as follows:
    {{du}\over{dx}}={{-R}\over{r\; x^2}}; hence, dx={{-r\; x^2}\over{R}} and  du={{-R}\over{r\; u^2}}\; du

    Now, you can substitute these into the integral as follows:
    \int {\rm atan}({{R}\over{r\; x}})\; dx={{-R}\over{r}}\int {{{\rm atan}(u)}\over{u^2}}\; du

    This integral can be found in a typical integral table.
    \int {{{\rm atan}(u)}\over{u^2}}\; du={{-1}\over{u}} \; {\rm atan}(u)\; - \; {{1}\over{2}} \; {\rm ln}\Biggl({{{1+u^2}\over{{u^2}}} \Biggr)\; +\; C, where  C is a constant.

    If you are required to solve this integral without using a table, then have a go at it. If you have trouble doing that, post back here and we can go into ways to solve it.

    If it turns out that the assumptions above are wrong, then please make the problem clear to us, and we can try again.
     
    Last edited: Dec 4, 2009
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Warning
    With the substitution u = R/rx the integral becomes improper if x = 0.

    Sometimes you can avoid the division by zero with a substitution, sometimes it makes things worse.
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    Are you sure about this? I have a doubt which I'll explain.

    The {\rm atan} function has an output range of -{{\pi}\over{2}} to {{\pi}\over{2}} for x equal to any real number. (note: I'm assuming he is not yet at a level of doing complex integration). In the limit as  x goes to zero, and  u goes to  \infty , the  {\rm atan}u function asymptotically approaches a constant level. Further, the actual function being integrated is {{{\rm atan}(u)\over{u^2}} which should then go to zero when  u goes to infinity. However, the decay looks to be  {{1}\over{u^2}} in the limit, which should converge when integrated.

    If there is any problem point, I would think they could be at  x=\pm \infty, where  u goes to zero. Here the numerator of the integration function (i.e.  {\rm atan}) has limiting form of  u and the denominator is  u^2 . This is a limiting  {{1}\over{u}} type dependence for the entire function, which is a singularity that won't converge when integrated.

    Please double check this and see if you agree.
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    I guess you did not have a chance to look at this yet. So I gave it some more thought. Thinking further, I should clarify the above comments a little bit more.

    The original integral that was given for atan(x) has no problem at x=0, but can't be integrated to infinity. This makes sense since atan goes to pi/2 at infinity (or -pi/2 at minus infinity). Integrating a constant value to infinity is not a converging case, obviously.

    The second integral that was given for atan(1/x) type functions has no problem at x=0 since that atan(1/x) function approaches a constant finite (i.e pi/2) value as x goes to zero, so the integral does converge to that limit. However, this integral also has problems with limits at infinity because the function looks like 1/x as x goes to infinity. We know that 1/x does not decay "fast" enough to be integrated to infinity.

    So basically, with the atan(1/x) type functions, plus and minus infinity are not limits that allow converging integrals. This means that a substitution need not be expected to handle those cases.

    The substitution I used above resulted in a new integral which also does converge if a limit is x=0 (u=infinity). However, that integral does not converge for the case of u=0 (x=infinity), which again corresponds with the original case.

    Hence, my view is that the substitution is working well in this problem.
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Steve you are absolutely right

    The simple arctan integral has a logarithmic term which of course becomes infinite when x=0.

    However the substitution u = R/rx leads to ln(1+0) when x=0.
    So this substitution does neatly sidestep that issue.

    Once again I was too hasty to work through it properly, although I blame the AAC snowflakes for affecting my aim.

    So it really does depend upon the nature of R and r in relation to x.
     
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