Question

I am trying to solve this integral:

Answer 1

\[\LARGE \int\limits_0^{\pi/4}e^{-a^2\sec^2(\theta)}d \theta\]

Answer 2

Do you think the antiderivative is an elementary function?

Answer 3

Not really, but suppose there is a nice form of it, call the integral this: \[I(a)\]

then the answer to this integral is: \[\large \int\limits_0^a e^{-x^2}dx=\sqrt{\frac{\pi}{4} -I(a)}\]

So it's probably just making the problem harder rather than easier. But I wouldn't mind getting an approximation to the erf function out of this if possible, I just hate not being able to calculate it.

Answer 4

\[\int\limits_0^{\pi/4} \sum\limits_{n=0}^{\infty}
\dfrac{(a\sec^2\theta)^n}{n!}d \theta \]

Answer 5

Good idea @ganeshie8 but don't forget the -1!

Answer 6

**\[\int\limits_0^{\pi/4} \sum\limits_{n=0}^{\infty}
\dfrac{(-1)^n(a \sec \theta)^{2n}}{n!}d \theta \]
?

Answer 7

but idk yet how to take this forward, lets continue your method :)

Answer 8

actually how about before that we use an identity to make it like this? I have no method anymore, this is where it ends basically haha. \[\LARGE \int\limits_0^{\pi/4}e^{-a^2\sec^2(\theta)}d \theta = \]
\[\LARGE e^{-a^2}\int\limits_0^{\pi/4}e^{-a^2\tan^2(\theta)}d \theta \]

Answer 9

Write the erf in its taylor series.

Answer 10

Nah, I don't want to do it that way.

Answer 11

oh sec^2 =tan^2+1?

Answer 12

have u thought about complexifying the cos

Answer 13

ur get e^f(e) and see i u can pull out e^something, to get it in that lambert W function form

Answer 14

random attempt lol

Answer 15

So currently it's kind of nailed down to this with J being the tangent version.

\[\large \int\limits_0^a e^{-x^2}dx=\sqrt{\frac{\pi}{4} -e^{-a^2}J(a)}\]

That's an interesting thought @dan815 idk I'm just doing random stuff.

Answer 16

what is this I(a) and J(a) stuff

Answer 17

oh here: \[\LARGE I(a)= \int\limits_0^{\pi/4}e^{-a^2\sec^2(\theta)}d \theta\] \[\LARGE J(a)= \int\limits_0^{\pi/4}e^{-a^2\tan^2(\theta)}d \theta\]

Answer 18

You can see in the limit as a goes to infinity that we get the correct answer, \[\int\limits_0^\infty e^{-x^2}=\frac{\sqrt{\pi}}{2}\] by looking at the term next to the J(a).

So it's not a bad approximation I guess

For anyone curious: http://www.wolframalpha.com/input/?i=d%2Fdx%28sqrt%28pi%29%2F2erf%28x%29%29

Answer 19

So my approximation then is: \[\large \frac{\sqrt{\pi}}{2}[erf(a)-erf(0)] \approx \sqrt{\frac{\pi}{4} - e^{-a^2}}\]

Answer 20

I'm trying to improve on it, since technically e^-a^2 is multiplied by a function of a, that integral J(a).

Answer 21

right okay i get it, its interesting

Answer 22

Actually I haven't at all explained how I got here to this, so I'll work on explaining that in a minute.