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How do i solve this integral? i used substitution but am not at all sure im right

  1. Apr 13, 2009 #1
    [tex]\int[/tex](tan(x))pdx (from 0 to pi/2)

    what i actually need to do is find the values of p so that the integral converges in its given limits
    i dont see too many options here, so what i did was
    t=tan(x)
    dt=t'dx=dx/cos2(x)=(1+tan2(x))dx=(1+t2)dx

    so now we have

    [tex]\int[/tex]tp(1+t2)dt (from 0 to infinity)

    and now i have no idea how to continue.
    is there a better way to solve this, if not how do i continue from here?
     
  2. jcsd
  3. Apr 13, 2009 #2
    How does t^p(1+t^2) behave near zero and for large t?
     
  4. Apr 13, 2009 #3
    i think that that would depend on p, would it not,
    for p>-2 it goes to infinity
    for p<2 it goes to 0

    but thats not the answer
     
  5. Apr 13, 2009 #4
    If you split the integral from zero to infinty to an integral from zero to R and an integral from R to infinity. Then for what p does the first integral converge and for what p does the second integral converge?
     
  6. Apr 13, 2009 #5
    for the 1st, when p>0
    for the 2nd, when p<-2
     
  7. Apr 13, 2009 #6
    I now see that you made a mistake in your first post. You have that:

    dx = dt/(1+t^2)

    So, the integral you need to consider is:

    t^p/(1+t^2)
     
  8. Apr 13, 2009 #7
    sorry, youre right,,,
    but that doesnt really make a difference to my inability so integrate this
     
  9. Apr 13, 2009 #8

    How does the function behave asymptotically for large t and for small t?
     
  10. Apr 13, 2009 #9
    for large t
    p<2 converges (0)
    p>2 diverges
    p=2 1

    for small t
    p>0 converges
    p<0 diverges
    p=0 1

    what are you trying to get me to say?...
    do i not need to actually integrate this to get the value of p??
     
  11. Apr 13, 2009 #10
    Does Integral of dt/sqr(t) from R to infinity exist?
     
  12. Apr 14, 2009 #11
    yes, but how do i get from t^p/(1+t^2)dt to 1/sqrt(t)dt
    i thought of changing u=t^p du=pt^p-1dt but that wont really help,
    to get to 1/sqrt(t)dt i could say u=(1+t^2)^2 du=2t(1+t^2)dt but i dont have that so what would that help.

    how can i solve this for P???!!!
     
  13. Apr 14, 2009 #12

    No, just forget about this particular problem for the moment and answer this question first: Suppose the problem is to fund out for which q the integral of t^(-q) dt from R to infinity converges. The answer is not q < 0.

    Similarly try to find out for which q the integral from zero to R of t^q dt converges.
     
  14. Apr 14, 2009 #13
    if i had to find q so that the integral of t^(-q) dt from R to infinity converges
    [tex]\int[/tex]t(-q) dt converges when q>1

    [tex]\int[/tex]tqdt
    =tq+1/(q+1)

    converges when q+1<0 therefore q<-1



    but i dont see how this is really relevant, what am i getting to by doing this?
     
  15. Apr 14, 2009 #14
    t^p/(1+t^2) is almost the same as t^q for some q for large t and it is almost t^q for a different q for small t.
     
  16. Apr 14, 2009 #15
    it seems like i have way too many unknowns to do anything here, i dont know what the q is in either case, dont know p,
     
  17. Apr 14, 2009 #16
    t^p/(1+t^2) = asymptotically t^(p-2) for large t, right?
     
  18. Apr 14, 2009 #17
    right,
    and would it be t^p for small t ?

    so would those be my q's???
    i still dont really see how this is going to solve my integral for p?
    could you please write out the steps, step by step to solving the problem,

    you have helped me each step but im not really seeing them as a whole problem, rather each step as its own
     
  19. Apr 14, 2009 #18
    tell me if this is right

    when t from R to infinity
    [tex]\int[/tex]tp/(1+t2)dt ~ [tex]\int[/tex]tp-2dt = [tex]\int[/tex]dt/t2-p
    2-p>1 ---> p<1 converges

    when t from 0 to R
    [tex]\int[/tex]tp/(1+t2) ~ [tex]\int[/tex]tpdt=[tex]\int[/tex]1/t-pdt
    -p<1 ---> p>-1 converges

    so now i know that the integral converges when p>-1 p<1, or in other words |P|<1
     
  20. Apr 14, 2009 #19
    but is this okay to do, using a function that is not exactly my function- i found when t^p-2 converges but that doesnt really mean that tp/1+t2 converges with the exact same p does it?
     
  21. Apr 14, 2009 #20
    Solving the integral is one thing, finding out for which p it converges is another thing. I'll show how to compte the integral after you find out for which p it converges.

    So far you have found that for small t you have that t^p/(1+t^2) is approximately t^p, while for large t it is approximately t^(p-2). Let's write the integral as:

    [tex]\int_{0}^{\infty}\frac{t^p}{1+t^2}[/tex] =

    [tex]\int_{0}^{\epsilon}\frac{t^p}{1+t^2} +\int_{\epsilon}^{R}\frac{t^p}{1+t^2} +\int_{R}^{\infty}\frac{t^p}{1+t^2} [/tex]

    For some small epsilon and large R. Then the middle term is finite, only the first and the last term are potentially divergent. You then investigate these two potentialy problemaic integrals separately.

    In case of the first integral, you can use that:

    t^p/(1+t^2) < t^p

    For p > -1 the integral of t^p fromzero to epsilon converges. So, this means that the integral of t^p/(1+t^2) will also converge for p>-1. But this doesn't prove that the integral of t^p/(1+t^2) doesn't converge for smaller p. All you know is that t^p won't converge if p is minus 1 or smaller, but t^p/(1+t^2) is smaller than t^p, so it doesn't follow that it won't converge.

    What you then do is prove that for the special case p = -1 the integral doesn't converge (this is trivial, just do a partial fraction expansion, you get a 1/t term which leads to a logarithmic divergence).

    Then for p < -1 you have that:

    t^p/(1+t^2) > t^(-1)/(1+t^2)

    if we choose epsilon smaller than 1. Then because the integral for p =-1 doesn't converge the intergal for p < -1 won't converge either.

    So, this then concludes the first integral.

    In case of the last integral, you use that:

    t^p/(1+t^2) = t^p/[t^2 (1+t^(-2))] =

    t^(p-2) 1/(1+t^(-2)) < t^(p-2)

    Then, you already obtained the result that the integral of t^(p-2) will converge for p-2<-1, so for p < 1. It doesn't converge for p = 1, but then since t^p/(1+t^2) < t^(p-2), this fact does not imply that our integral does not converge for p = 1. To show that it indeed does not converge for p = 1, you need to compute the integral of t/(1+t^2) from t = R to infiinty and show that it diverges. Then you use the fact that for p > 1 you have:

    t^p/(t+t^2) > t/(1+t^2)

    provided R was chosen larger than 1.

    So, the final result is that the first integral converges for p > -1, while thelast integral converges for p < 1, therefore the integral from zero to infinity converges when -1<p<1.
     
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