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I am trying to find the constant for the second excited wave state that will normalize it. The equation for the second excited state is $$\psi_2(x)=A_2(1-\frac{2m\omega x^2}{\hbar})e^{-\frac{mwx^2}{2\hbar}}$$ therefore I am trying to solve $$1=\int_{-\infty}^{+\infty}A_2^2(1-\frac{2m\omega x^2}{\hbar})^2e^{-\frac{mw x^2}{\hbar}}dx$$ which can be simplified to $$1=2A_2^2\int_{0}^{\infty}(1-\frac{2m\omega x^2}{\hbar}^2)e^{-\frac{mw x^2}{\hbar}}dx$$ Which will give three integrals to solve so it can be expressed as $$1=2A_2^2(I_1+I_2+I_3)$$ where $$I_1=\int_{0}^{\infty}e^{-\frac{mw x^2}{\hbar}}dx \\ I_2=-4\int_{0}^{\infty}e^{-\frac{3mw x^2}{2\hbar}}dx \\ I_3=4\int_0^{\infty}e^{-\frac{2m\omega x^2}{\hbar}}dx $$

I'm wondering if there's a fairly quick method to evaluate these three integrals other than using say WolframAlpha

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To solve $I_1=\int_{0}^{\infty}e^{-\frac{mw x^2}{\hbar}}dx$, let $k = \frac{mw}{h}$, then you just need to solve $I_1=\int_{0}^{\infty}e^{kx^2}dx$. This integral comes up in one of the problems in Griffiths Quantum Mechanics, and solutions are easy to find online. If I recall correctly, Griffiths references using a table of known integrals to solve it.

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There is a classic trick for integrating gaussians $\int_0^\infty e^{-x^2} dx$.

$$\int_0^\infty e^{-x^2} dx=\sqrt{\int_0^\infty e^{-x^2} dx*\int_0^\infty e^{-y^2} dy}=\sqrt{\int_0^\infty \int_0^\infty e^{-x^2-y^2} dx dy}$$

Then switch to polar coordinates $r^2=x^2+y^2$ and $dxdy=rdrd\theta$ and noticing that the integral is over a quarter of the plane so that $$\sqrt{\int_0^\infty \int_0^\infty e^{-x^2-y^2} dx dy}=\sqrt{\int_0^\infty \int_0^{\frac{\pi}{2}} e^{-r^2} rdr d\theta}$$ and then its a u substiton $u = r^2$, $\frac{du}{2} = r dr$ and the $\theta$ integral is easy.

$$\sqrt{\frac{\pi}{2}\int_0^\infty e^{-u} \frac{du}{2}}=\sqrt{\frac{\pi}{4}*1}=\frac{\sqrt{\pi}}{2}.$$

The rest is just making the proper substitions for your $I_1,I_2,I_3$ integrals.

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