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




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((1x)^2) will go to zero also just as fast as exp((0x)^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 MeijerG 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.



