I am attempting to compute a quantum mechanics problem, considering a particle in a harmonic potential. I have simplified the probability relation to: $$P=\frac{2}{\sqrt{\pi}}\int_1^{\infty}e^{-{u^2}}du,$$ My problem is computing this integral
My attempt
I know that the Euler–Poisson integral equates to: $$\int_{-\infty}^{\infty}e^{-u^2}du=\sqrt{\pi},$$ I was initially thinking perhaps, due to the symmetry of the Gaussian function. I could consider my problem as: $$P=\sqrt{\pi}\int_{-(\infty+1)}^{\infty}e^{-{u^2}}du,$$ which is 'essentially the same' as: $$P=\sqrt{\pi}\int_{-\infty}^{\infty}e^{-{u^2}}du,$$ and hence: $$P=\pi.$$ Although this seems logical and gives a nice answer, I seem to have found a contradiction to my solution:
By the same logic, surely I can consider my integral: $$P=\frac{2}{\sqrt{\pi}}\int_1^{\infty}e^{-{u^2}}du=\frac{1}{\sqrt{\pi}}\left[\int_{1}^{\infty}e^{-{u^2}}du+\int_{-\infty}^{-1}e^{-{u^2}}du\right],$$ Surely though: $$\left[\int_{1}^{\infty}e^{-{u^2}}du+\int_{-\infty}^{-1}e^{-{u^2}}du\right]=\left[\int_{-\infty}^{\infty}e^{-{u^2}}du-\int_{-1}^{1}e^{-{u^2}}du\right],$$ I know that for $u=0$ the integral results in $1$, thus: $$\eta=\int_{-1}^{1}e^{-{u^2}}du>1,$$ but if this is correct, then surely: $$P=\frac{1}{\sqrt{\pi}}\left[\sqrt{\pi}-\eta\right],$$ where $\eta>1.$ But the only way for this to agree with my initial solution is if:$$\eta=(\sqrt{\pi}-\pi\sqrt{\pi})$$ which makes no sense, as $\eta$ cannot be negative
Problem
I'm am definitely incorrect somewhere in one of these attempts, otherwise they would agree, I am thinking that it has something to do with me assuming that $\infty=\infty+1,$ (in the first part) but am posting here to see if it is even possible to compute this integral by hand to get an exact solution, and if it is how to do so. Thanks in advance.