$\int\limits_{-\infty}^{\infty}(f(x))^{a} dx$ , $0 < a < 1$ given $\int\limits_{-\infty}^{\infty}f(x) dx$

Asked Dec 23 '16 at 22:57

Active Dec 23 '16 at 22:57

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For any $f(x)$, are there any tricks to compute $\int\limits_{-\infty}^{\infty}(f(x))^{a} dx$ , $0 < a < 1$

given a known $\int\limits_{-\infty}^{\infty}f(x) dx$ ?

I thought differentiation under integral sign might lead me to something like this, but I couldn't get much further with that.

asked Dec 23 '16 at 22:57

Srini

Contour integration seems very usable here. Though I can't think up any general method. – Simply Beautiful Art Dec 23 '16 at 23:01
It is not the case that the integral also exists, given some $f(x)$ and $a\in(0,1)$. – Simply Beautiful Art Dec 23 '16 at 23:05
I don't really think this is possible. Consider $f=\chi_{(0,1)}$ and $f=2\chi_{(0,1/2)}$ and $\alpha=1/2$. Both have the same integral but when you take a square root you get two different results. – Dec 23 '16 at 23:06
Indeed the latter does not determine the former. – Did Dec 23 '16 at 23:12

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