Why does substituting u = t² in the Gaussian integral give zero?

[Heresy42]

Dear All,

I computed an integral that looks like erf(x) without problem: $$\int_{-\infty}^{+\infty} e^{-t^2} dt = \int_{-\infty}^{0} e^{-t^2} dt + \int_{0}^{+\infty} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} [-erf(-\infty)+erf(+\infty)] = \frac{\sqrt{\pi}}{2} [-(-1)+1] = \sqrt{\pi}.$$

However, what about the change of variable: $$u = t^2?$$
Hence: $$du = 2 dt$$ and: $$\int_{-\infty}^{+\infty} e^{-t^2} dt = \frac{1}{2} \int_{+\infty}^{+\infty} e^{-u} du = -\frac{1}{2} [e^{-u}]_{+\infty}^{+\infty} = 0.$$

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.

 

[Whovian]

This is exactly why one splits up an integral from -∞ to ∞ into two parts and takes the limit of each: the normal rules of integration don't necessarily apply.

 

[daveb]

However, what about the change of variable: $$u = t^2?$$
Hence: $$du = 2 dt$$

No

du=2tdt

 

[Heresy42]

Indeed, my bad, thanks to both of you.

 

[D H]

However, what about the change of variable: $$u = t^2?$$
Hence: $$du = 2 dt$$ and: $$\int_{-\infty}^{+\infty} e^{-t^2} dt = \frac{1}{2} \int_{+\infty}^{+\infty} e^{-u} du = -\frac{1}{2} [e^{-u}]_{+\infty}^{+\infty} = 0.$$

You've already been told about the problem with du. Another problem here: u cannot be negative, yet you have the integrations limits being -∞ to +∞.

 

[Heresy42]

Dear D H,

Thanks for your reply. It's now clear to me that this change of variable is for positive values only.

Regards.