To approximate e^(-x) for large x in integrals, one can utilize the property that e^(-x) can be expressed as e^(-x-x0)*e^(x0), allowing for simplifications. In the context of evaluating the integral for the probability of an electron being more than 1 meter away from the nucleus, the rapid decay of the exponential means that contributions from regions where r > 1m are negligible. The integral can be simplified to a form that is easy to evaluate, yielding an approximate value of 10^(-10^10). Additionally, methods like Gauss–Laguerre quadrature can be employed for numerical integration, but an analytical solution is also achievable. Ultimately, the discussion highlights both approximation techniques and the potential for exact solutions in specific cases.
#1
Chain
35
3
I'm trying to evaluate an integral with e^-x where x is huge in the domain of the integral so I can't evaluate it numerically without making an approximation.
Can you add more context? There might be a clever way to approximate e^(-x), but if that value is not added to something, e^(-x)=e^(-x-x0)*e^(x0) where the second factor is independent of x and the first factor can be chosen to be about e^0.
#3
Chain
35
3
The integral is 4*(r^2)*exp(-2*r/a)/a^3 integrated between 1 and ∞ (The probability an electron in the ground state of hydrogen is more than 1 metre away from the nucleus) a=0.529*10^-10
Note that your integrand is one you can anti-differentiate without much trouble, so you can get the exact answer. (e.g. integration by parts. Or computer algebra package)
If you just want an estimate: The exponential will drop very quick, so regions with r>1m+eps are irrelevant and r^2 is nearly constant and =1m^2. Therefore, the integral is simply 4m^2/a^3 * exp(-2r/a) which can be evaluated as 2m^2/a^2 * exp(-2m/a) ≈ 8*10^20 * exp(-4*10^10) ≈ 10^(-10^10) where the last approximation is very rough.
Fair enough, yeah I realized after posting this the integral could be solved analytically >__< and I got a valule of something like 10^(-10^10) but thanks for the responses :)