- #1
Heresy42
- 5
- 0
Dear All,
I computed an integral that looks like erf(x) without problem: [tex] \int_{-\infty}^{+\infty} e^{-t^2} dt = \int_{-\infty}^{0} e^{-t^2} dt + \int_{0}^{+\infty} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} [-erf(-\infty)+erf(+\infty)] = \frac{\sqrt{\pi}}{2} [-(-1)+1] = \sqrt{\pi}. [/tex]
However, what about the change of variable: [tex] u = t^2? [/tex]
Hence: [tex] du = 2 dt [/tex] and: [tex] \int_{-\infty}^{+\infty} e^{-t^2} dt = \frac{1}{2} \int_{+\infty}^{+\infty} e^{-u} du = -\frac{1}{2} [e^{-u}]_{+\infty}^{+\infty} = 0. [/tex]
I just want to be sure: the second integration doesn't make sense, because the lower and upper bounds are the same, right?
Thanks.
Regards.
I computed an integral that looks like erf(x) without problem: [tex] \int_{-\infty}^{+\infty} e^{-t^2} dt = \int_{-\infty}^{0} e^{-t^2} dt + \int_{0}^{+\infty} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} [-erf(-\infty)+erf(+\infty)] = \frac{\sqrt{\pi}}{2} [-(-1)+1] = \sqrt{\pi}. [/tex]
However, what about the change of variable: [tex] u = t^2? [/tex]
Hence: [tex] du = 2 dt [/tex] and: [tex] \int_{-\infty}^{+\infty} e^{-t^2} dt = \frac{1}{2} \int_{+\infty}^{+\infty} e^{-u} du = -\frac{1}{2} [e^{-u}]_{+\infty}^{+\infty} = 0. [/tex]
I just want to be sure: the second integration doesn't make sense, because the lower and upper bounds are the same, right?
Thanks.
Regards.