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Test the improper integral ∫ e-x2 for convergence

Test the following improper integral for convergence:

    \[ \int_{-\infty}^{\infty} e^{-x^2} \, dx. \]


The integral converges.

Proof. We have

    \begin{align*}  \int_{-\infty}^{\infty} e^{-x^2} \, dx &= \int_{-\infty}^0 e^{-x^2} \, dx + \int_0^{\infty} e^{-x^2} \, dx \\[9pt]  &= 2 \int_0^{\infty} e^{-x^2} \, dx &(e^{-x^2} \text{ is an even function})\\[9pt]  &= 2 \left( \int_0^1 e^{-x^2} \, dx + \int_1^{\infty} e^{-x^2} \, dx \right). \end{align*}

The first integral converges since it is a proper integral. For the second integral, since x \geq 1 we have e^{-x^2} \leq e^{-x}. But, we know \int_1^{\infty} e^{-x} \, dx converges (by Example #4 on page 418 of Apostol with s = 0). Hence, we have established the convergence of

    \[ \int_{-\infty}^{\infty} e^{-x^2} \, dx. \qquad \blacksquare \]

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