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Let $f:[0,\infty) \rightarrow \mathbb R$ be a strictly positive, decreasing, differentiable function, such that $$f(0) = 1, \quad \lim_{x\rightarrow \infty} f(x) = 0$$ and $$\frac{1}{f(x)^2} = \frac{1}{f(x^2)} + 2x^2$$ If $\int_0^\infty f(x)\,dx$ exists, show that $$\int_0^\infty f(x^2) \,dx = \frac{1}{\sqrt2}\int_0^\infty f(x)\,dx$$

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Is this homework? Did you try anything? Maybe a change of variables $u=x^2$? –  yohBS Mar 21 '12 at 19:23
    
Hmm, it's not at all obvious to me how to proceed... from the information given I'm tempted to try something like writing both sides as functions of a second variable and doing some differentiation. +1, +star, and looking forward to seeing the answer. –  user7530 Mar 21 '12 at 21:43
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$f(x)=\frac{1}{2x}$ :) –  no identity Mar 21 '12 at 21:50
    
Martin, how did you come across this problem? I wouldn't be surprised if you had a means of constructing these types of identities... –  Antonio Vargas Mar 22 '12 at 18:08
    
Thank you Antonio for your answer. It was just a simple observation that the function $\frac{1}{1 + x^2}$ satisfies the functional equation that you solved. –  Martin Mar 23 '12 at 17:31

1 Answer 1

For my answer I need to assume that $f(x)$ is analytic at $x=0$ and meromorphic on $\mathbb{C}$.

We will focus on the functional equation

$$ \frac{1}{f(x)^2} = \frac{1}{f(x^2)} + 2x^2, \tag{1} $$

which may be rearranged to

$$ f(x^2) = f(x)^2 + 2x^2 f(x^2) f(x)^2. \tag{1*} $$

Since $f$ is analytic at $x=0$ we can write

$$ f(x) = \sum_{n=0}^{\infty} a_n x^n $$

near $x=0$. Substituting this into $(1^*)$ we get

$$ \sum_{n=0}^{\infty} a_n x^{2n} = \sum_{n=0}^{\infty} b_n x^n + 2 x^2 \sum_{n=0}^{\infty} c_n x^n, \tag{2} $$

where

$$ b_n = \sum_{\genfrac{}{}{0pt}{1}{p+q=n}{p,q \geq 0}} a_p a_q = \sum_{k=0}^{n} a_k a_{n-k} \hspace{1.3cm} \text{and} \hspace{1.3cm} c_n = \sum_{\genfrac{}{}{0pt}{1}{2p+q=n}{p,q \geq 0}} a_p b_q = \sum_{k=0}^{\lfloor n/2 \rfloor} a_k b_{n-2k}. \tag{3} $$

Before we go any further, let us rewrite the sums in $(2)$ to make equating coefficients easier. We get

$$ a_0 + \sum_{n=1}^{\infty} a_n x^{2n} = b_0 + b_1 x + \sum_{n=2}^{\infty} b_n x^n + \sum_{n=2}^{\infty} 2 c_{n-2} x^n. \tag{2*} $$

We immediately get $a_0 = b_0 = a_0^2$, so that $a_0 = 1$. We also immediately get $0 = b_1 = 2a_0 a_1$, so that $a_1 = 0$.

We will show that $a_n = 0$ for all odd $n$. Indeed, for $n \geq 3$ odd we have from $(2^*)$ and $(3)$ that

$$ 0 = b_n + 2c_{n-2} = b_n + 2\sum_{k=0}^{(n-3)/2} a_k b_{n-2-2k}. $$

If we suppose that $b_k = 0$ for all odd $k < n$ (which is at least true when $k=1$), the sum on the right, involving only $b$'s with odd index, vanishes. Thus we have $b_n = 0$. By induction it follows that $b_n = 0$ for all odd $n$, and hence from the definition of $b_n$ in $(3)$ we may conclude, again by induction, that $a_n = 0$ for all odd $n$, as claimed.

We may use this fact to obtain new formulas for $b_{2n}$ and $c_{2n}$. We get

$$ b_{2n} = \sum_{k=0}^{n} a_{2k} a_{2n-2k} \hspace{1.3cm} \text{and} \hspace{1.3cm} c_{2n} = \sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} b_{2n-4k}. \tag{4} $$

We now consider even $n$. For all $n \geq 2$ even we have from $(2^*)$ that

$$ a_{n/2} = b_n + 2c_{n-2}. \tag{5} $$

As a taste of what's ahead, let's substitute $n=2$ in $(5)$. Since $a_1 = 0$ we get

$$ 0 = b_2 + 2c_0 = 2a_2 + 2, $$

so that $a_2 = -1$.

We will show that $a_{2n} = (-1)^n$ for all $n \geq 0$. Suppose $a_{2k} = (-1)^k$ for all $k \leq n$ (which is at least true for $k=0,1$), and replace $n$ with $2n+2$ in $(5)$ to get $a_{n+1} = b_{2n+2} + 2c_{2n}$, which becomes, with the help of $(4)$,

$$ \begin{align} \frac{1-(-1)^n}{2} (-1)^{(n+1)/2} &= 2a_{2n+2} + \sum_{k=1}^{n} a_{2k} a_{2n+2-2k} + 2c_{2n} \\ &= 2a_{2n+2} + \sum_{k=1}^{n} (-1)^{k+n+1-k} + 2c_{2n} \\ &= 2a_{2n+2} - (-1)^n n + 2c_{2n} \\ &= 2a_{2n+2} - (-1)^n n + 2\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} b_{2n-4k} \\ &= 2a_{2n+2} - (-1)^n n + 2\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} \sum_{\ell=0}^{n-2k} a_{2\ell} a_{2n-4k-2\ell} \\ &= 2a_{2n+2} - (-1)^n n + 2\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} \sum_{\ell=0}^{n-2k} (-1)^{\ell+n-2k-\ell} \\ &= 2a_{2n+2} - (-1)^n n + 2\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} (-1)^n (n-2k+1) \\ &= 2a_{2n+2} - (-1)^n n + 2(-1)^n (n+1) \sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} - 4(-1)^n \sum_{k=0}^{\lfloor n/2 \rfloor} k\, a_{2k} \\ &= 2a_{2n+2} - (-1)^n n + 2 (-1)^n (n+1) \frac{1+(-1)^{\lfloor n/2 \rfloor}}{2} \\ &\hspace{6cm} - 4(-1)^{n+\lfloor n/2 \rfloor} \left\lceil \frac{\lfloor n/2 \rfloor}{2} \right\rceil. \end{align} $$

The identity

$$ \begin{align} 2(-1)^{n+1} &= \frac{1-(-1)^n}{2} (-1)^{(n+1)/2} + (-1)^n n - 2 (-1)^n (n+1) \frac{1+(-1)^{\lfloor n/2 \rfloor}}{2} \\ &\hspace{8cm} + 4(-1)^{n+\lfloor n/2 \rfloor} \left\lceil \frac{\lfloor n/2 \rfloor}{2} \right\rceil \end{align} $$

can be checked by testing the possible behaviors of $n$ mod 4, by which we obtain $a_{2n+2} = (-1)^{n+1}$. This completes the inductive step.

Thus

$$ f(x) = 1 - x^2 + x^4 - x^6 + x^8 + \cdots $$

near zero. We assumed that $f$ is meromorphic on $\mathbb{C}$, and $f$ has an analytic continuation to $\mathbb{C} \setminus \{-i,i\}$ by the formula $f(x) = 1/(1+x^2)$, so we must have $f(x) = 1/(1+x^2)$ on $\mathbb{C}$.

We note that $f$, as we've just defined it, satisfies $f(0) = 1$ and $\lim_{x \to \infty} f(x) = 0$, and $f$ is decreasing on $[0,\infty)$.

It thus remains to note that

$$ \int_{0}^{\infty} f(x^2)\,dx = \int_{0}^{\infty} \frac{dx}{1+x^4} = \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{dx}{1+x^2} = \frac{1}{\sqrt{2}}\int_{0}^{\infty} f(x) dx, $$

which completes the proof.

(Many thanks to my officemate for some valuable discussion on this :) )

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There is a typo at the end—in the last line you should have the integral of $(1+x^2)^{-1}$ (hence of $f(x)$) instead of the integral of $(1+x^2)^{-2}$. –  Nick Strehlke Mar 22 '12 at 19:13
    
Thanks for pointing that out, @Nick. –  Antonio Vargas Mar 22 '12 at 19:41
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No problem; nice argument by the way. –  Nick Strehlke Mar 22 '12 at 20:04
    
@Nick, thank you! –  Antonio Vargas Mar 22 '12 at 21:07

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