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Math Help - Help, Urgent and difficult (and entertaining) integral

  1. #1
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    Question Help, Urgent and difficult (and entertaining) integral

    Forum-Wanderers I am in great need of your help, I am currently researching a project in financial mathematics with deadline today. I have achieved a number of thing but one small problem eludes me. Please help me finish with a bang by helpping me solve


    P(x)= - < integral between pi/4 and pi/2 of > [ (1/2pi)*exp(-sect/2)dt ] + 1/8


    I have tried to make it look something like the formula for the probability of t beeing in a given interval [a;b] under a normally istributed law:


    = - < integral between a and b of > [ (1/sqrt(2pi))*exp(t/2)dt ]


    I have already simplified it from the earlier expressio, for the curious:


    < integral between 0 and 1 of > {(1/sqrt(2pi))*exp(x/2)*<integral between x and + oo of > ((1/sqrt(2pi))*exp(y/2)dy)dx}


    In advance, many thanks,
    I hope to get replies soon
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  2. #2
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by carolus4 View Post
    Forum-Wanderers I am in great need of your help, I am currently researching a project in financial mathematics with deadline today. I have achieved a number of thing but one small problem eludes me. Please help me finish with a bang by helpping me solve


    P(x)= - < integral between pi/4 and pi/2 of > [ (1/2pi)*exp(-sect/2)dt ] + 1/8


    I have tried to make it look something like the formula for the probability of t beeing in a given interval [a;b] under a normally istributed law:


    = - < integral between a and b of > [ (1/sqrt(2pi))*exp(t/2)dt ]


    I have already simplified it from the earlier expressio, for the curious:


    < integral between 0 and 1 of > {(1/sqrt(2pi))*exp(x/2)*<integral between x and + oo of > ((1/sqrt(2pi))*exp(y/2)dy)dx}


    In advance, many thanks,
    I hope to get replies soon
    \int_0^1 \left\{\frac{1}{\sqrt{2\pi}}\exp\left(\frac{x^2}{2  }\right) \int_x^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left(\fra  c{y^2}{2}\right) \, dy\right\}\, dx
     = \frac{1}{2\pi} \int_0^1\int_x^{\infty}\exp\left(\frac{x^2}{2}\rig  ht) \cdot \, \exp\left(\frac{y^2}{2}\right) \, dy \, dx = \frac{1}{2\pi} \int_0^1\int_x^{\infty}\exp\left(\frac{x^2+y^2}{2}  \right) \, dy \, dx

    from here, you can do the integral by transforming it to polar coordinates..
    let x = r\cos\theta and y = r\sin\theta

    can you continue from here?
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  3. #3
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    well I have, and that gave me the first integral.

    Am I mistaken? (please say yes!! at least that means my problem is solvable)

    I did take a while to see the new limits were now

    theta between pi/2 and pi/4
    r between 0 and 1/cos(r)

    (I got the cos r result from simple trig of the area limited by
    x between 0 and 1
    y above y=x)


    this still leaves me stumped with a sec I'd like to be rid of.


    thanks for your reply
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