# X=^-e^{t} integral

eljose
let be the integral:

$$\int_0^{\infty}F(x)dx$$ with $$x\rightarrow\infty F(x)\rightarrow0$$ then we make the change of variable x=-e^{t} then the new integral would become $$\int_{-\infty}^{\infty+i\pi}F(-e^t)e^{t}dt$$ my question is if we can ignore the integral from $$(\infty,\infty+i\pi)$$ so we have only the integral $$\int_{-\infty}^{\infty}F(-e^t)e^{t}dt$$ as for big value the F(x) tends to 0

Last edited:

Homework Helper
You cannot make this substitution. Note that x > 0, and et is also always greater than 0, so how could x = -et ever be true?

eljose
why not? the same would happen with x=-1/(t+1) x>0 but t<0

Last edited:
Homework Helper
How can a negative quantity equal a positive quantity?

quetzalcoatl9
i think that you lost a negative sign when making your substitution.

the example you give in post #3 is correct for x>0 t<-1

so it sounds like the unresolved question is your bounds on the integral. I'm not sure how to handle this, but keep two things in mind:

1) remember that you are dealing with a formal limit

$$\lim_{a \rightarrow -\infty} \int_{a}^{i\pi} f(x) dx + \lim_{b \rightarrow \infty} \int_{i\pi}^{b} f(x) dx$$

2) you are integrating over $$\mathbb{C}$$ so you need to get a book on complex integration and see what to do. i suspect that solving the above and integrating as usual will suffice.

eljose
Another question let,s suppose we have the integral:

$$\int_{-\infty}^{\infty}F(x)dx$$ and make the change of variable x=t+ai with i
=sqrt(-1) then what would be the new limits?..thanks...

$$\int _{-\infty} ^{\infty} F(x)dx$$