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Homework Statement
[tex]
\int_{0}^{\pi/4} \exp \left(-\frac{1}{\cos^2x}\right)\mbox{d}x
[/tex]
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...
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.The value of the integral, whatever it is, is not going to have a simple form: The "[URL [Broken] value[/URL] is not recognized by http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&number=.227651878&lookup_type=browse".
Edit: I think I overestimated the capabilities of Plouffe's inverter, so maybe the solution is simple afterall.
[tex]Write the integral as exp(-a*sec(x)^2). Find d/da of that.
hmmm...I don't see how because:Now THAT function you can integrate dx from 0 to pi/4 (in terms of erf)..
This isn't the most easy integral I tackledNot quite. The argument of the erf is t. t isn't equal to tan(x). Now keep going.
[tex]You are getting there. There's also a 2/sqrt(pi) in the erf definition. What happened to that?
Of course! The derative of erf(v) is e^{-v2 }!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?
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.Of course! The derative of erf(v) is e^{-v2 }!
Right, I'll try again: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?
Right, but for all [itex]a[/itex] you just showedIf a = 0 then
[tex]
\int_{0}^{\pi/4} \exp \left(-\frac{a}{ \cos^2x}\right)\mbox{d}x = \frac{ \pi}{4}
[/tex]
Not a very useful conclusion; try solving for [itex]C_0[/itex] instead.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]
Not at all. Reread gabbagabbahey's post and pay attention this time. You know what erf(0) is.So [tex]C_0 =0[/tex]
It agreesSure. I did a numerical integration and I got 0.22765187804641. Does that agree with your result?