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This question is motivated by my previous post about sinc function.

Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire (i.e. analytic everywhere) function $f(x)$ on $\mathbb{R}$ such that $$\int_0^\infty f(x) dx=\int_0^\infty f(x)^2 dx$$ or $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx.$$

If $f$ is required only to be continuous, then other examples are possible, e.g. the even extension of the following function: $$ f(x)=\left\{\begin{array}{ll} -2(5+\sqrt{65})x^2+(7+\sqrt{65})x-1 & 0\le x\le \frac{1}{2}\\ 2(5+\sqrt{65})x^2-(13+3\sqrt{65})x+4+\sqrt{65} & \frac{1}{2}\le x\le 1\\ \frac{1}{x^2} & x\ge 1 \end{array}\right. $$

As commented below, it turns out that there are easy answers to the above question. AD also showed a function below that also satisfies $$ \sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty f(n)^2=0. $$ In view of these answers, my question is now revised to:

Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire function, $f(x)$ on $\mathbb{R}$ such that $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx=\sum_{-\infty}^\infty f(n) =\sum_{-\infty}^\infty f(n)^2 $$

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Take any analytic $f(x)$ with nonzero integral. Solve $\int af(x)dx = \int (af(x))^2dx$ for $a$. – Zarrax Dec 8 '10 at 19:17
$\int_0^\infty {2e^{ - ax} dx} = \int_0^\infty {4e^{ - 2ax} dx} = 2/a$, for any $a > 0$. – Shai Covo Dec 8 '10 at 19:27
How about f(x)=0? – Ross Millikan Dec 8 '10 at 21:33
SEe the word "nonzero"? – TCL Dec 8 '10 at 21:34

Similar to the suggestion of Zaricuse, that is take an entire function $f$ such that $\int_{-\infty}^\infty f(x) dx \ne0$ then solve for $$\int_{-\infty}^\infty af(x)dx = \int_{-\infty}^\infty (af(x))^2dx$$ Then $g(z)=af(z)$ solves half of the problem. To reach $$\sum g(n)=\sum g(n)^2$$ we may for example start with $f(z)=\sin (\pi z) \cdot h(z)$ where $h$ is an other integrable entire function.

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Less clumsy than mine. – Ross Millikan Dec 8 '10 at 22:11
@Ross Millikan: Maybe, but it is a bit cheap. – AD. Dec 9 '10 at 6:00

In the spirit of Zaricuse's solution without the sum requirement, take any sufficiently well behaved functions f and g. Then you should be able to find a linear combination af+bg that satisfies both equations. If you let

$$if=\int_{-\infty}^\infty f(x) dx$$ $$if2=\int_{-\infty}^\infty f(x)^2 dx$$ $$sf=\sum_{n=1}^\infty f(n)$$ $$sf2=\sum_{n=1}^\infty f(n)^2$$

and similarly for fg and g, we have

$$a*if + b*ig=a^2*if2+2ab*ifg+b^2ig2$$ and
$$a*sf + b*sg=a^2*sf2+2ab*sfg+b^2sg2$$

which can be solved for a and b in most cases.

Added In response to the new request that the values of the two integrals and two sums all match, I just need enough knobs to turn. Define $g(k,x)=\exp(-kx^2)$ and take $f(x)=g(1,x)+ag(2,x)+bg(3,x)+cg(4,x)$ The nice thing about this $f$ is that $f^2$ is written in terms of $g(k,x)$, though k goes up to 8. The integral of $g(k,x)$ is just $\sqrt{\frac{\pi}{k}}$ and the sum is calculated by Wolfram Alpha as $\vartheta_3(0,\exp(-k))$. We can make a table:

$$\begin{array}{ccc}k&\int g(k,x)&\sum g(k,x)\\1&1.772453851&1.77264\\2&1.253314137&1.27134\\3&1.023326708&1.09959\\4&0.886226925&1.03663\\5&0.79266546&1.01348\\6&0.723601255&1.00496\\7&0.669924586&1.00182\\8&0.626657069&1.00067\end{array}$$

So the integral of f is $\sqrt{\pi}(1+a/\sqrt{2}+b/\sqrt{3}+c/\sqrt{4})$ The integral of f^2 is $\sqrt{\pi}(1/\sqrt{2}+2a/\sqrt{3}+(a^2+2b)/\sqrt{4}+(2c+2ab)/\sqrt{5}+(b^2+2ac)/\sqrt{6}+2bc/\sqrt{7}+c^2/\sqrt{8})$ with similar expressions for the sum in terms of theta3. We want to find a,b,c so that the integrals and sums all match. Unless my matrix of coefficients has a very unlikely dependence it will be available. Excel claims $a=-3.782590725, b=4.503400057, c=-1.83137936$ is very close to a solution, and there should be more.

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Of course, if you can get the exact values of a,b,c, or you can prove that such solutions for a,b,c exist, then your solution will be perfect. – TCL Dec 9 '10 at 19:33
@TCL: I doubt it. Theta functions don't really admit closed-form inverses. – J. M. Dec 10 '10 at 1:25
@J.M. I don't need inverses, just values. Then I get three equations in the three unknowns a, b, and c, two quadratic and one linear. There has to be a solution, but without exact values for theta3 (and is sqrt(pi) an exact value?) it's hard to demonstrate. I suppose one might use the implicit function theorem to prove existence. – Ross Millikan Dec 10 '10 at 2:38
Then I misread the whole thing, Ross. If you do have "three equations in the three unknowns a, b, and c, two quadratic and one linear", then Gröbner should prove adequate in finding the needed values. – J. M. Dec 10 '10 at 2:43
Mathematica 's NSolve[] returns four solutions to this problem. I did not have the patience to wait for the Solve[] version, but from the looks of things the exact results are too complicated to be easily manipulable. – J. M. Dec 10 '10 at 3:56
up vote 1 down vote accepted

The revised question has been answered in this post at MO. In particular, it was shown that $\frac{\sin ax}{ax}$ satisfies this equality for each $0<a\le \pi$.

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