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$$\int e^{-x^2} dx$$ How do we calculate this integration?

نبع الطيب
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    ... this function does not have an antiderivative given in terms of what are called elementary functions. See here http://math.stackexchange.com/questions/265780/how-to-determine-with-certainty-that-a-function-has-no-elementary-antiderivative and http://math.stackexchange.com/questions/239105/special-integrals If you REALLY want a formula, an antiderivative of $e^{-x^2}$ is given by $F(x)=\int_a^x e^{-t^2},dt$. – JP McCarthy Oct 09 '14 at 16:21
  • An other question would be not to look for the anti-derivative, but for the result of $\int_{-\infty}^{+\infty} e^{-x²}dx$ – Martigan Oct 09 '14 at 16:28

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Using the series expansion \begin{align} e^{-x^{2}} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{2n}}{n!} \end{align} then the integral has the form \begin{align} \int e^{- x^{2}} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{2n+1}}{n! \, (2n+1)}. \end{align} The series presented here is defined as the series for the error function and the integral is given by \begin{align} \int e^{- x^{2}} \, dx = \frac{\sqrt{\pi}}{2} \, erf(x). \end{align}

Leucippus
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We can calculate the value of integral over $\mathbb{R}$ as follows:

Let $A=\int_{-\infty}^{\infty} e^{-x^2} \, dx=\int_{-\infty}^{\infty} e^{-y^2} \, dy$.

$A^2=\int_{-\infty}^{\infty} e^{-x^2} \, dx\int_{-\infty}^{\infty} e^{-y^2} \, dy=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} \, dxdy$ (since $x,y$ are independent). Then you can change to polar coordinate to evaluate the double integral. Since $e^{-x^2}>0$, we take square root to get the final result.

John
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