- #1
cozycoz
I want to integrate this:[tex] \int_0^∞ re^{-\frac{1}{2σ^2} (r-iσ^2q)^2} \, dr. [/tex]
If I change the variable r into t with this relation:[tex]r-iσ^2q=t,[/tex]
then the integral becomes[tex]\int_{-iσ^2q}^∞ (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt[/tex]
so it seems I cannot use the famous gaussian integral formula. But I got the correct result for the imaginary part of this problem with just ignoring the lower limit(actually I just forgot to rearrange upper and lower limits but anyway):[tex]iσ^2q\int_{0}^{∞} e^{-\frac{1}{2σ^2} t^2}\,dt= \frac{iσ^2q}{2} \int_{-∞}^∞ e^{-\frac{1}{2σ^2}t^2} \, dt =\frac{iσ^2q}{2}\sqrt{2σ^2π}[/tex] So why are they the same: the integration over the interval {##[-iσ^2q, 0]+[0, ∞]##} and ##[0, ∞]##?
If I change the variable r into t with this relation:[tex]r-iσ^2q=t,[/tex]
then the integral becomes[tex]\int_{-iσ^2q}^∞ (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt[/tex]
so it seems I cannot use the famous gaussian integral formula. But I got the correct result for the imaginary part of this problem with just ignoring the lower limit(actually I just forgot to rearrange upper and lower limits but anyway):[tex]iσ^2q\int_{0}^{∞} e^{-\frac{1}{2σ^2} t^2}\,dt= \frac{iσ^2q}{2} \int_{-∞}^∞ e^{-\frac{1}{2σ^2}t^2} \, dt =\frac{iσ^2q}{2}\sqrt{2σ^2π}[/tex] So why are they the same: the integration over the interval {##[-iσ^2q, 0]+[0, ∞]##} and ##[0, ∞]##?