21
$\begingroup$

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 $$\int_0^\infty f(x^2) \,dx = \frac{1}{\sqrt2}\int_0^\infty f(x)\,dx$$

$\endgroup$
5
  • $\begingroup$ Is this homework? Did you try anything? Maybe a change of variables $u=x^2$? $\endgroup$
    – yohBS
    Mar 21 '12 at 19:23
  • $\begingroup$ 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. $\endgroup$
    – user7530
    Mar 21 '12 at 21:43
  • 1
    $\begingroup$ $f(x)=\frac{1}{2x}$ :) $\endgroup$
    – Norbert
    Mar 21 '12 at 21:50
  • $\begingroup$ Martin, how did you come across this problem? I wouldn't be surprised if you had a means of constructing these types of identities... $\endgroup$ Mar 22 '12 at 18:08
  • $\begingroup$ 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. $\endgroup$
    – Martin
    Mar 23 '12 at 17:31
15
$\begingroup$

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 :) )

$\endgroup$
3
  • $\begingroup$ 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 $(1+x^2)^{-2}$. $\endgroup$ Mar 22 '12 at 19:13
  • $\begingroup$ Thanks for pointing that out, @Nick. $\endgroup$ Mar 22 '12 at 19:41
  • 1
    $\begingroup$ No problem; nice argument by the way. $\endgroup$ Mar 22 '12 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.