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How do I solve for $A$ such that $\int_{-\infty }^{\infty } A e^{-\lambda(x-a)^2}\, dx = 1$?

I first attempt u-substitution like so:

Let $u = x-a$

$\implies du = dx$ and $x = u + a$, so

$$ \int_{-\infty }^{\infty } A e^{-\lambda(x-a)^2}\, dx = \int_{-\infty }^{\infty } A e^{-\lambda u^2}\, du $$

After this point I am stuck. Do I need to do another u-substitution?

user1068636
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    Try $t=\sqrt{\lambda} u$ and recalling the Gaussian integral. – Vishu Jul 15 '21 at 14:20
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    $ \displaystyle I = \int_{-\infty }^{\infty } A e^{-\lambda u^2}, du = \int_{-\infty }^{\infty } A e^{-\lambda v^2}, dv.$ So $ \displaystyle I^2 = \int_{-\infty }^{\infty } \int_{-\infty }^{\infty } A^2 e^{-\lambda (u^2+v^2)}, du \ dv$. Now convert to polar coordinates and evaluate $I^2$. – Math Lover Jul 15 '21 at 14:41
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    Just to make it explicit: the Guassian integral is the integral $\displaystyle\int_{-\infty}^\infty e^{-x^2}dx$. – A-Level Student Jul 15 '21 at 17:11

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$$I=\int_{-\infty}^\infty A\exp(-\lambda(x-a)^2)dx=\int_{-\infty}^\infty A\exp(-(\sqrt\lambda x)^2)dx$$ then say $u=\sqrt\lambda x\Rightarrow dx=du/\sqrt\lambda$ and so you arrive at: $$\frac{A}{\sqrt\lambda}\underbrace{\int_{-\infty}^\infty e^{-u^2}du}_{\sqrt\pi}=1$$ Now you should be able to solve it from here

Henry Lee
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