TITLE: Integral for function of square
QUESTION [21 upvotes]: 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
$$\int_0^\infty f(x^2) \,dx = \frac{1}{\sqrt2}\int_0^\infty f(x)\,dx$$

REPLY [15 votes]: 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 :) )