Normalized primitive of an exponential flat factor

Statement

Fix $c>0$ , an integer $j\ge0$ , and a smooth function $b(y,\eta)$ for $|y|<\varepsilon$ , $\eta\in U\subseteq\mathbb R^d$ . For $0<y<\varepsilon$ , define

H(y,\eta)=e^{c/y^2}\int_0^y e^{-c/u^2}u^j b(u,\eta)\,du.

Then $H(y,\eta)=y^{j+3}B(y,\eta)$ , where $B$ extends smoothly through $y=0$ and

B(0,\eta)=\frac{b(0,\eta)}{2c}.

Fixed-domain integral proof

Use the substitution $s=u^{-2}-y^{-2}$ . It gives

B(y,\eta)=\frac12\int_0^\infty e^{-cs} (1+y^2s)^{-(j+3)/2} b\!\left(\frac{y}{\sqrt{1+y^2s}},\eta\right)\,ds.

The right side is defined for both signs of small $y$ . On compact parameter sets, every derivative of its integrand is bounded by $C(1+s)^N e^{-cs}$ , for some $C,N$ depending on that derivative. This is integrable, so differentiation under the integral proves smoothness. Set $y=0$ to obtain the stated boundary value.

Positive smooth multipliers

If $a(y)=e^{-c/y^2}g(y)$ for $y>0$ , with $g$ smooth and $g(0)>0$ , then

\frac1{a(y)}\int_0^y a(u)u^j b(u,\eta)\,du

also equals $y^{j+3}$ times a smooth coefficient near zero. Apply the formula with $gb$ and divide the resulting coefficient by the nonvanishing function $g(y)$ . This conclusion uses the specific exponential profile, not just flatness.