I have a numerical evidence of $$\int_0^{1/2} \frac{1}{\sqrt{2\pi}\sigma_0x}\exp\left(-\frac{(\mu_0x-y)^2}{2\sigma_0^2x^2}\right)dx \approx 1+\cos(2\pi y),$$ where $$\sigma_0^2=\frac{2}{\pi^2}-\mu_0^2$$ and $$\mu_0=\frac{4}{\pi^2}.$$ The latter should be the mean and the variance of the function $$h(x)= \pi(1-x)\sin(\pi(1-x)).$$ Could anybody suggest a proof turning the numerical approximate equality above into an equality?
Proof of a gaussian integral turning into a cosine
2
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calculus
normal-distribution
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1Equality is dubious. If you are after an approximation, in which regime for $y$? – 2012-12-03
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0$y\in(0,1/2),$ the same range as $x$. Numerically (integration step $\Delta x=10^{-4}$) the two curves are really superimposed, so to make me suspect a formal equality. Thanks. – 2012-12-03
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0Your suspicion is not justified, try $y=0$. – 2012-12-03
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0Yes, for $y=0$ it does not work but already for $y=5\cdot 10^{-4}$ the error is less than 2%. – 2012-12-03
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0So we are not talking about formal equality, right? Regarding approximate numerical agreement, I am a little suspicious about it when y is small since, when y goes to zero, the LHS goes to infinity while the RHS goes to 2. Likewise, I suspect the LHS goes to zero when y goes to infinity while the RHS does not. All this brings us back to my first question: which regime for y? – 2012-12-03
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0Again, $y$ is between 0 and 1/2. In my original problem, $y$ is the normal-distributed variable, while its variance and mean scale with a parameter $x$. Now, integrating over $x$ gives the result that I was honestly expecting. I have calculated $\sigma_0$ and $\mu_0$ from a function that actually reads $$h(y,x)=\frac{\pi(y)}{x^2}\sin(\pi\frac{y}{x})$$ for $x=1$. Now I have chosen $h(x,y)$ because $\int_{y}^{1/2}h(x,y)dx=1+cos(2\pi y)$. – 2012-12-03
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0Did you try another integration algorithm. It is quite suspicious that you get something oscillatory out of this integral. – 2012-12-03
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0Actually there is no complete oscillation since the period is twice the integration domain. The resulting integral approximated by $1+cos(2\pi y)$ is more like an S-function thus it could have something to do with $erf$'s or similar. – 2012-12-03