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Using Integral Calculator, the error function is used for the indefinite integral, and then $\operatorname{erf}(\infty)=1$ and $\operatorname{erf}(\infty)=-1$ gives the answer $\frac{\sqrt{\pi}}{\sqrt{2a}}$. Is there another way to solve the definite integral, or do you need to use the error function to solve this?

Lorenzo B.
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oswinso
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    look up "Gaussian integral". This was topic discussed around $10^{23123^7}$ times only here on MSE.... – tired May 27 '18 at 11:19

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You can use that $$\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$$ now substitute $u^2=2ax^2$, and then $du=\sqrt{2a}dx$, so $$\int_{-\infty}^\infty e^{-2ax^2}dx=\frac{1}{\sqrt{2a}}\int_{-\infty}^\infty e^{-u^2}du=\frac{1}{\sqrt{2a}}\sqrt{\pi}$$