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Determine whether the integral is convergent or divergent. $\int^{\infty}_{-\infty} 7xe^{-x^2}dx$

It equals zero. Yet, I am unsure if it is convergent or divergent. Can someone tell me and explain why?

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  • $\begingroup$ An integral from $\infty$ to $\infty$ does not make much sense. Did you mean $-\infty$ to $\infty$? $\endgroup$ – Ian Oct 25 '15 at 17:49
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    $\begingroup$ Once you know it converges, it is equal to 0 because it is an odd function over a symmetric interval. Convergence as per below $\endgroup$ – Alan Oct 25 '15 at 17:53
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    $\begingroup$ A definite integral is convergent if and only if it's equal to a real number. In this case, it's equal to $0$, therefore it's convergent. $\endgroup$ – user236182 Oct 25 '15 at 17:53
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    $\begingroup$ @user236182 "A definite integral is convergent if and only if it's equal to a real number.": completely meaningless sentence. $\endgroup$ – Najib Idrissi Oct 25 '15 at 17:54
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    $\begingroup$ Until you know that the integral is convergent, it's too early to say that it equals zero. If it's divergent, it doesn't equal anything; it's simply just divergent. $\endgroup$ – Hans Lundmark Oct 25 '15 at 18:13
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HINT:

The indefinite integral

$$\int 7xe^{-x^2}\,dx=-\frac72 e^{-x^2}+C$$

and the limit

$$\lim_{x\to \pm \infty }e^{-x^2}=0$$

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