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$$\frac{1}{\sqrt{2\pi s^2}} \int_{-\infty}^{\infty} xe^{-(x-m)^2/(2s^2)} \, dx$$

I am stuck at this problem. Please give some hint as how to initiate?

Mathematics
  • 3,973

2 Answers2

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Hint: Make the change of variable $t = \frac{x - m}{s}$, and use the celebrated Gaussian integral formula: $$\int_{-\infty}^\infty e^{-\frac{t^2}{2}} dt = \sqrt{2\pi}. \tag{$*$}$$

For a proof for $(*)$, check this link. Also, the first answer to this math stack exchange post referred a paper that gives $10$ different ways to prove $(*)$, which is very fun to read.

If you have learned some probability, you may identify this expression as the mean value of normal distribution $\mathcal{N}(m, s^2)$, which is $m$.

Zhanxiong
  • 14,040
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Hint: $$\int_{-\infty}^\infty xe^{-(x-m)^2/(2s^2)}\,dx =\int_{-\infty}^\infty (x-m)e^{-(x-m)^2/(2s^2)}\,dx +m\int_{-\infty}^\infty e^{-(x-m)^2/(2s^2)}\,dx\ .$$

David
  • 82,662