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I don't know if this is a very simple question with a very difficult answer, but:

Is $y = \dfrac{\sin x}{x}$ the only function such that

$\int_{-\infty}^{\infty} f(x) dx =\int_{-\infty}^{\infty} f^2(x) dx \text{ ?}$

This is to say, if we start with the integral equation

$\int_{-\infty}^{\infty} f(x) dx =\int_{-\infty}^{\infty} f^2(x) dx \text{ ?}$

would we only find $f(x) = \dfrac{\sin x}{x}$, or we can expect other solutions?

Maybe I should have made this clear, but I'm talking about $f(x) \neq 0$ and $f(x)$ continuous in $\mathbb{R}$.

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    @HenningMakholm Oh, I mean $(f(x))^2$. For $f(f(x))$, I usually say "$f$ of $f$", or $f \circ f(x)$2012-03-02

3 Answers 3

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Take any $f$ where both integrals exist. Define $ A = \int_{-\infty}^{\infty} f(x) dx, $ then $ B = \int_{-\infty}^{\infty} f^2(x) dx. $ Now, define $ g(x) = \frac{A}{B} \; f(x). $ Then $ \int_{-\infty}^{\infty} g(x) dx = \int_{-\infty}^{\infty} g^2(x) dx = \frac{A^2}{B}. $

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    @Peter: Yes, indeed; the set of functions to which you can do what Will and I show is the intersection $L^1\cap L^2$.2012-03-02
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Note that $ \int_{-\infty}^\infty\frac{1}{1+x^2}\mathrm{d}x=\pi $ and $ \int_{-\infty}^\infty\frac{1}{(1+x^2)^2}\mathrm{d}x=\frac{\pi}{2} $ However, if we scale this up, we get $ \int_{-\infty}^\infty\frac{2}{1+x^2}\mathrm{d}x=2\pi $ and $ \int_{-\infty}^\infty\frac{4}{(1+x^2)^2}\mathrm{d}x=2\pi $ This can be done for any function so that $f$ and $f^2$ are integrable and $\int_{-\infty}^\infty f(x)\:\mathrm{d}x\not=0$.

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    Could you help me out [here](http://math.stac$k$exchange.com/questions/107715/the-definition-of-the-logarithm)?2012-03-02
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Not by far.

In fact, take any nonnegative continuous $g$ such that $\int g^2 > \int g$. Then select some $x_0$ such that $0 and consider the functions $ f_a(x) = \begin{cases}g(x) & x\in(-\infty,x_0] \\ g(x_0) & x\in(x_0,x_0+a] \\ g(x-a) & x \in (x_0+a,\infty) \end{cases}$ for all $a>0$. As you increase $a$, both of $\int f_a^2$ and $\int f_a$ will increase in proportion to $a$, but $\int f_a$ will increase faster by a factor of $1/g(x_0)$. Since $f_0=g$, at some $a$ you will find $\int f_a^2=\int f_a$.

Alternatively, take the same $g$ and let $b=\frac{\int g}{\int g^2}$ Then $f(x)=bg(x)$ will also satisfy $\int f=\int f^2$.

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    I like your alternative :-) (+1)2012-03-02