# Thread: Help, Urgent and difficult (and entertaining) integral

1. ## 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(-sec²t/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(t²/2)dt ]

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

< integral between 0 and 1 of > {(1/sqrt(2pi))*exp(x²/2)*<integral between x and + oo of > ((1/sqrt(2pi))*exp(y²/2)dy)dx}

I hope to get replies soon

2. Originally Posted by carolus4
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(-sec²t/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(t²/2)dt ]

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

< integral between 0 and 1 of > {(1/sqrt(2pi))*exp(x²/2)*<integral between x and + oo of > ((1/sqrt(2pi))*exp(y²/2)dy)dx}

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?

3. 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 sec² I'd like to be rid of.