1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hard integral

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \int_{0}^{\pi/4} \exp \left(-\frac{1}{\cos^2x}\right)\mbox{d}x
    [/tex]



    3. The attempt at a solution
    Can someone give me a hint? I don't see a smart substitution nor a path via integration by parts...
     
  2. jcsd
  3. Mar 1, 2009 #2
    Last edited by a moderator: May 4, 2017
  4. Mar 1, 2009 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are right. Plouffe's inverter is not all powerful. It only knows about numbers it's happened to have heard about. But there is a reasonably simple expression. You do need the erf function. Write the integral as exp(-a*sec(x)^2). Find d/da of that. Now THAT function you can integrate dx from 0 to pi/4 (in terms of erf). You can in turn integrate that expression da. And you know the value if a=0. It's not terribly simple, to find the constant of integration you need to figure out things like lim a->0 erf(sqrt(a))/sqrt(a). But it can be done. You might want to put (pi/4)*(1-erf(1)^2) into Plouffe's inverter.
     
    Last edited by a moderator: May 4, 2017
  5. Mar 2, 2009 #4
    [tex]

    \frac{d}{da}\ \left[ \int_{0}^{\pi/4} \exp \left(-a \cdot \sec^2 (x) \right) \mbox{d}x \right ] = \int_{0}^{\pi/4} - \sec^2(x) \cdot \exp \left(-a \sec^2 x\ \right) \mbox{d}x

    [/tex]

    hmmm...I don't see how because:

    [tex] \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt. [/tex]
     
  6. Mar 2, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Be creative! You have to be. This whole problem is about trickyness. sec(x)^2=1+tan(x)^2. Now substitute t=sqrt(a)*tan(x). You know there is an answer. I sketched the route. I even gave you a really strong hint what the answer was. Try and figure out one or two of the non-obvious steps without the need for a hint.
     
    Last edited: Mar 2, 2009
  7. Mar 2, 2009 #6
    [tex]
    \frac{dI}{da}=\int_{0}^{\pi/4} - \sec^2(x) \cdot \exp \left(-a \sec^2 x\ \right) \mbox{d}x = ( -e^{-a}) \cdot \int_0^{\pi /4} \sec^2(x) \cdot \exp (-a \tan^2 x )\ \mbox{dx}

    [/tex]


    Now with

    [tex] t =\sqrt{a} \tan(x) \longrightarrow t^2 = a \tan^2(x) [/tex]

    and

    [tex] dt = \sqrt{a} \sec^2(x) [/tex]



    I get:

    [tex] \frac{dI}{da} = \frac{-e^{-a}}{ \sqrt{ a} } \cdot \int_0^{ \pi /4} e^{-t^2} \mbox{d}t =\frac{-e^{-a}}{ \sqrt{ a} } \cdot \mbox{erf} ( \pi /4) [/tex]
     
  8. Mar 2, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Not quite. The argument of the erf is t. t isn't equal to tan(x). Now keep going.
     
  9. Mar 2, 2009 #8
    This isn't the most easy integral I tackled :rolleyes:

    Ok, so the boundaries are wrong (that's what you meant in your last post, right Dick?)


    [tex]
    \frac{dI}{da} = \frac{-e^{-a}}{ \sqrt{ a} } \cdot \int_0^{ \sqrt{a}} e^{-t^2} \mbox{d}t =\frac{-e^{-a}}{ \sqrt{ a} } \cdot \mbox{erf} ( \sqrt{a})
    [/tex]

    right?
     
  10. Mar 2, 2009 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are getting there. There's also a 2/sqrt(pi) in the erf definition. What happened to that?
     
  11. Mar 2, 2009 #10
    [tex]

    \frac{dI}{da} = \frac{-e^{-a}}{ \sqrt{ a} } \cdot \int_0^{ \sqrt{a}} e^{-t^2} \mbox{d}t = \left( \frac{ \sqrt{ \pi}}{2} \right) \left( \frac{-e^{-a}}{ \sqrt{ a} } \right) \cdot \mbox{erf} ( \sqrt{a})

    [/tex]
     
  12. Mar 2, 2009 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Good. Now can you integrate that da? It's creative time. It's not NEARLY as hard as it looks. Think u substitution.
     
  13. Mar 3, 2009 #12
    Now with [tex]u=\sqrt{a}[/tex]

    I get:

    [tex] \int \left( \frac{ \sqrt{ \pi}}{2} \right) \left( \frac{-e^{-a}}{ \sqrt{ a} } \right) \cdot \mbox{erf} ( \sqrt{a}) \mbox{d}a =\int (-\sqrt{ \pi}) \cdot e^{-u^2} \cdot \mbox{erf} (u)\ \mbox{d}u = C \cdot \int \int e^{u^2-t^2}\ \mbox{d}t \mbox{d}u
    [/tex]

    This is getting nasty...
     
  14. Mar 3, 2009 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If u=erf(sqrt(a)), what's du? Or since you've already done that u substitution for sqrt(a), if v=erf(u), what's dv?
     
  15. Mar 3, 2009 #14
    Of course! The derative of erf(v) is e-v2 !

    But then

    [tex] \int \left( \frac{ \sqrt{ \pi}}{2} \right) \left( \frac{-e^{-a}}{ \sqrt{ a} } \right) \cdot \mbox{erf} ( \sqrt{a}) \mbox{d}a =\int (-\sqrt{ \pi}) \cdot e^{-u^2} \cdot \mbox{erf} (u)\ \mbox{d}u = C_1 \cdot \mbox{erf} (u) + C_2

    [/tex]


    With the very important integration constant C2, right?
     
  16. Mar 3, 2009 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Dead right, but don't forget the sqrt(pi) and 2 parts of the erf definition. That's why it's actually an easy integral. But doing the v=erf(u) substitution turns the integral into v*dv, right? What's that? And there's no 'integration constant' C1. You know what C1 is. C2 is important, yes.
     
  17. Mar 4, 2009 #16
    Right, I'll try again:

    [tex]
    \int \left( \frac{ \sqrt{ \pi}}{2} \right) \left( \frac{-e^{-a}}{ \sqrt{ a} } \right) \cdot \mbox{erf} ( \sqrt{a}) \mbox{d}a =\int (-\sqrt{ \pi}) \cdot e^{-u^2} \cdot \mbox{erf} (u)\ \mbox{d}u = (\sqrt{ \pi}) \frac{ \sqrt{\pi} }{2} \int v\ \mbox{d}v = \frac{1}{4} \pi v^2 +C_0
    [/tex]


    I previously used C_1 to 'hide' all constants but I didn't specify that, C_0 is here the constant of integration multiplied by some factor. Is this correct, Dick?
     
  18. Mar 4, 2009 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You forgot the minus sign. So you've got -pi*erf(sqrt(a))^2/4+C. Now how to determine C?
     
  19. Mar 7, 2009 #18
    If a = 0 then


    [tex]

    \int_{0}^{\pi/4} \exp \left(-\frac{a}{ \cos^2x}\right)\mbox{d}x = \frac{ \pi}{4}

    [/tex]



    So I conclude that:

    [tex] \frac{\pi}{4} = \left( \frac{- \sqrt{\pi}}{2} \right) \cdot \lim_{a \rightarrow 0} \int e^{-a} \cdot \frac{ \mbox{erf} (\sqrt{a}) }{\sqrt{a}}\ \mbox{d}a [/tex]
     
  20. Mar 7, 2009 #19

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Right, but for all [itex]a[/itex] you just showed

    [tex]\int_{0}^{\pi/4} \exp \left(-\frac{a}{ \cos^2x}\right)\mbox{d}x = \frac{1}{4} \pi \mbox{erf}(\sqrt{a})^2 +C_0[/tex]


    Not a very useful conclusion; try solving for [itex]C_0[/itex] instead.

    What is [tex] \frac{1}{4} \pi \mbox{erf}(\sqrt{a})^2 +C_0[/tex] for a=0?
     
  21. Mar 7, 2009 #20
    So [tex]C_0 =0[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hard integral
  1. Hard integral (Replies: 6)

  2. Hard Integral (Replies: 5)

  3. Hard integrals (Replies: 25)

  4. Hard integral (Replies: 5)

  5. Hard Integral (Replies: 8)

Loading...