integral e^(2x^2) Sorry; it's been a while.
S spoon3457 New member Joined Sep 27, 2006 Messages 3 Sep 27, 2006 #1 integral e^(2x^2) Sorry; it's been a while.
pka Elite Member Joined Jan 29, 2005 Messages 11,982 Sep 27, 2006 #2 There is no elementary anti-derivative for that.
S spoon3457 New member Joined Sep 27, 2006 Messages 3 Sep 27, 2006 #3 why not 3/4e^(2x^2) its my integrading fator for a differential quation
S spoon3457 New member Joined Sep 27, 2006 Messages 3 Sep 27, 2006 #4 oops the origonal problem was e^(4x^2)
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 27, 2006 #5 These types of 'e' integrals are difficult to integrate. As pka said, not by elementary means, i.e. parts, u-substitution, etc. This can be done, but by other means, such as asymptotic series, error function, complex analysis, etc. Anyway, I ran it through Maple and here's what she gave me: \(\displaystyle \L\\\int{e^{4x^{2}}}dx=\frac{1}{4}\frac{\sqrt{\pi}erf(2\sqrt{-ln(e)})}{\sqrt{-ln(e)}}\) I don't know why it displayed −ln(e)\displaystyle \sqrt{-ln(e)}−ln(e) instead of −1=i\displaystyle \sqrt{-1}=i−1=i erf=error function
These types of 'e' integrals are difficult to integrate. As pka said, not by elementary means, i.e. parts, u-substitution, etc. This can be done, but by other means, such as asymptotic series, error function, complex analysis, etc. Anyway, I ran it through Maple and here's what she gave me: \(\displaystyle \L\\\int{e^{4x^{2}}}dx=\frac{1}{4}\frac{\sqrt{\pi}erf(2\sqrt{-ln(e)})}{\sqrt{-ln(e)}}\) I don't know why it displayed −ln(e)\displaystyle \sqrt{-ln(e)}−ln(e) instead of −1=i\displaystyle \sqrt{-1}=i−1=i erf=error function