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The integral in question is $\int_{-\infty}^{+\infty} x {e}^{-\lambda x^2}dx$ where $x$ and $\lambda$ both are real numbers.

My solution:

$\int_{-\infty}^{+\infty} x {e}^{-\lambda x^2}dx = \int_{-\infty}^{0} x {e}^{-\lambda x^2}dx +\int_{0}^{+\infty} x {e}^{-\lambda x^2}dx =$
$\lim_{a\rightarrow -\infty}\int_{a}^{0} x {e}^{-\lambda x^2}dx +\lim_{b\rightarrow +\infty}\int_{0}^{b} x {e}^{-\lambda x^2}dx = \begin{vmatrix} u = -\lambda x^2\\ du = -\lambda 2xdx\\ \begin{matrix} x & 0 & a & b\\ u & 0 & -\lambda a^2 & -\lambda b^2 \end{matrix} \end{vmatrix} =$
$-\frac{1}{2\lambda}\left( \lim_{a\rightarrow -\infty}\int_{-\lambda a^2}^{0} {e}^{u}du +\lim_{b\rightarrow +\infty}\int_{0}^{-\lambda b^2} {e}^{u}du \right) =$
$-\frac{1}{2\lambda}\left( \lim_{a\rightarrow -\infty}\left(1 - {e}^{-\lambda a^2} \right) +\lim_{b\rightarrow +\infty}\left({e}^{-\lambda b^2} - 1 \right) \right) = -\frac{1}{2\lambda}\left( \left(1 - 0 \right) +\left(0 - 1 \right) \right) = 0$

1) Is this solution correct?
2) Suppose that real function of real argument $f\left(x\right)$ is odd and both limits of $f\left(x\right)$ as $x$ approaches $\pm\infty$ are finite values. Is it enough to say that $\int_{-\infty}^{+\infty}f\left(x\right)dx$ is equal to 0?

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I think you need $\lambda > 0$ for the integral to converge (in which case it is zero) – Juan S Mar 27 '11 at 22:33
up vote 4 down vote accepted
  1. Yes.

  2. This only works if $f$ is integrable. A counterexample would be $f(x)=x/(1+x^2)$, whose integral doesn't exist. What you need to know is that $\int_a^b f(x)dx\to 0$ as $a$ and $b$ go to infinity, and this will hold if $f$ is integrable.

However, you could say that the Cauchy principal value of the integral is $0$ when $f$ is odd.

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