Substitution Rule for Infinite Limits of Integration

[ehrenfest]

[SOLVED] limits of integration

Homework Statement

I want to substitute x=t^2 in \int_{-\infty}^{\infty}{\exp(-t^4)} dt. What are the new limits of integration? They are both infinity aren't they? But the integral is clearly not zero? Is the problem that the substitution rule only holds for finite limits of integration?

Homework Equations

The Attempt at a Solution

 

[Mathdope]

The substitution introduces dx = 2t dt within the integral. Perhaps a polar coordinate form would be helpful.

 

[ehrenfest]

Of course, but 2t = 2 \sqrt{x}. Why is this not valid:

\int_{-\infty}^{\infty}\exp(-t^4)dt = \int_{\infty}^{\infty}\exp(-x^2)dx/(2\sqrt{x})

?

 

[NateTG]

Well, you need to use -\sqrt{-x} where x<0.

 

[Mathdope]

Maybe you should break the integral up into the part over t < 0 and t > 0. What does that turn into? I don't know the answer, just a suggestion.

 

[ehrenfest]

I don't care how to evaluate the specific integral. My question is about the substitution rule. No one has yet explained to me why this

\int_{-\infty}^{\infty}\exp(-t^4)dt = \int_{\infty}^{\infty}\exp(-x^2)dx/(2\sqrt{x})

is not a valid use of the substitution rule?

 

[Mystic998]

I think it's just because t^2 = x doesn't necessarily imply that t = sqrt(x).