Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Topic: integration question: 1/(Sqrt[1 - x^2])*Exp[-(1 - x)^2]
Replies: 4   Last Post: Aug 5, 2013 5:51 PM

 Messages: [ Previous | Next ]
 clicliclic@freenet.de Posts: 1,245 Registered: 4/26/08
Re: integration question: 1/(Sqrt[1 - x^2])*Exp[-(1 - x)^2]
Posted: Aug 5, 2013 5:51 PM

"Nasser M. Abbasi" schrieb:
>
> Integration experts:
>
> Looking at
>
> 1/(Sqrt[1 - x^2])*Exp[-(1 - x)^2]
>
> and trying to integrate it from zero to infinity.
>
> When replacing Exp[-(1 - x)^2] by Exp[-(x)^2] it works.
> For example, on Mathematica:
>
> ------------------------
> Integrate[1/(Sqrt[1 - x^2])*Exp[-(x)^2], {x, 0, Infinity}]
> (* -((I*BesselK[0, -(1/2)])/(2*Sqrt[E])) *)
> -------------------------
>
> Notice:
>
> Limit[1/(Sqrt[1 - x^2])*Exp[-(x^2)], x -> Infinity]
>
> gives zero as expected. But
>
> Limit[1/(Sqrt[1 - x^2])*Exp[-(1 - x^2)], x -> Infinity]
>
> gives
>
> DirectedInfinity[-I]
>
> So the problem is that replacing Exp[-(x)^2] by Exp[-(1 - x)^2]
> makes the integrand blow up.
>
> And I am not sure I understand this part since as x->large value, then
> exp(-(1-x)^2) will go to zero also just as fast as exp(-(0-x)^2)?
>

The pole of 1/(Sqrt[1 - x^2] at x = 1 is integrable, and the numerical
values provided by Mathermatica and Maple look approximately correct.
Apparently, your original integrand cannot be handled symbolically via
Meijer-G convolution, whereas the modified integrand 1/(Sqrt[1 - x^2])*
Exp[-(x)^2] can.

Your replacing Exp[-(x^2)] by Exp[-(1 - x^2)] = Exp[-1 + x^2] rather
than Exp[-(1 - x)^2] causes the integrand to blow up as x tends to
infinity.

Martin.

Date Subject Author
8/5/13 Nasser Abbasi
8/5/13 Axel Vogt
8/5/13 Nasser Abbasi
8/5/13 Nasser Abbasi
8/5/13 clicliclic@freenet.de