# Integral formula for $f(\sqrt x)$, where $f$ is smooth and even

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be even smooth function. Let $w(x)=f(\sqrt x)$ for $x>0$. On MO http://mathoverflow.net/questions/65264 I found the following integral formula $$w^{(k)}(x^2)=\frac{(2x)^{-2k+1}}{(k-1)!} \int_0^x (x^2-t^2)^{k-1} f^{(2k)}(t) dt.$$

How to proof it and show that there exist limits $\lim_{x\rightarrow 0} w^{(k)}(x^2)$ for $k=1,2,\ldots$? By it would be follow smoothness of $w$ on $[0,\infty)$.

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Make the substitution $t=xy$. We get \begin{align*} w^{(k)}(x^2)&=\frac{(2x)^{-2k+1}}{(k-1)!}\int_0^1(x^2-x^2y^2)^{k-1}f^{(2k)}(xy)xdy\\ &=\frac{2^{-2k+1}}{(k-1)!}x^{-2k+1}x^{2(k-1)}x\int_0^1(1-y^2)^{k-1}f^{(2k)}(xy)dy\\ &=\frac{2^{-2k+1}}{(k-1)!}\int_0^1(1-y^2)^{k-1}f^{(2k)}(xy)dy, \end{align*} and by the dominated convergence theorem $$\lim_{x\to 0}\: w^{(k)}(x^2)=f^{(2k)}(0)\int_0^1(1-y^2)^{k-1}dy.$$ You can get the integral formula by induction: show it for $k=1$ and if it's true for $k$ then $$2xw^{(k+1)}(x^2)=\frac{2^{-2k+1}}{(k-1)!}\int_0^1(1-y^2)^{k-1}f^{(2k+1)}(xy)ydy,$$ and integrating by parts \begin{align*} w^{(k+1)}(x^2)&=\frac{2^{-2k-1}}{x(k-1)!}\int_0^12y(1-y^2)^{k-1}f^{(2k+1)}(xy)dy \\ &=\frac{2^{-2k-1}}{x(k-1)!}\left(\left[-\frac{(1-y^2)^k}kf^{(2k+1)}(xy)\right]_{y=0}^{y=1}+\frac 1k\int_0^1(1-y^2)^kxf^{(2k+2)}(xy)dy\right)\\ &=\frac{2^{-2(k+1)+1}}{k!}\int_0^1(1-y^2)^kf^{(2(k+1))}(xy)dy. \end{align*} (since $f^{(2k+1)}(0)=0$ for all $k\geq 1$).