What would the new limits be for this integral?

[Xyius]

I am working on an physics problem and it has boiled down to this integral.

$$\int_{0}^{∞} r e^{-\frac{1}{2 r_0}(r-i r_0^2 q)^2}dr$$

I found that if I make the substitution $u=r-i r_0^2 q$, then I can do the integration, but I am a little confused about what the limits would be in terms of u. Normally I would just replace u with its expression in terms of r, but one of the integrals turns out being a gaussian integral and thus, doesn't have an indefinite form.

I have for the upper limit to be ∞, but the lower limit is $-i r_0^2 q$. This lower limit obviously will not work for the gaussian integral. Plus it doesn't make sense to me, a negative, complex radial limit?

 

[Simon Bridge]

I am working on an physics problem and it has boiled down to this integral.

$$\int_{0}^{∞} r e^{-\frac{1}{2 r_0}(r-i r_0^2 q)^2}dr$$

I found that if I make the substitution $u=r-i r_0^2 q$, then I can do the integration, but I am a little confused about what the limits would be in terms of u. Normally I would just replace u with its expression in terms of r, but one of the integrals turns out being a gaussian integral and thus, doesn't have an indefinite form.

$$\int_a^b f(x)dx = \int_{u(a)}^{u(b)}g(u)du$$

I have for the upper limit to be ∞, but the lower limit is $-i r_0^2 q$. This lower limit obviously will not work for the gaussian integral. Plus it doesn't make sense to me, a negative, complex radial limit?

I see... perhaps the way to make sense of it for you is to separate the integrand into real and imaginary parts before you do the substitution.

You can also consider that u is not a radius.