Assume $f(x)$ is a real, differentiable and continuous function over $R$, I want to calculate an integration : $$ \lim_{a\to \infty} \int_0^a f^m(x) \frac{df(a-x)}{d(a-x)}dx. $$ where $m$ is any positive integer. The conditions are: $f(x) = f(-x)$, $f(0) = 1$, $\frac{df}{dx}|_{x=0} = -1$, and $f(\infty)$ is also known, defined as $f_{\infty}$. I tried in this way: $$ \begin{align} &\int_0^a f^m(x) \frac{df(a-x)}{d(a-x)}dx \\ &= -\int_0^a f^m(x) \frac{df(a-x)}{dx}dx \\ &= -\frac{d}{dx}\int_0^a f^m(x) f(a-x) dx \\ \end{align} $$ Here I use the relation that the derivation is distributive over the convolution. I am also not sure that whether the calculation is correct. Then I can not go any further.
Could someone give me some suggestions on this problem please?
$$f^m(x)\frac{df(a-x)}{d(a-x)}dx\tag{1}$$ notice that: $$\frac{d(a-x)}{dx}=-1$$ and so: $$\frac{1}{d(a-x)}=-\frac 1{dx}$$ which gives: $$(1)=f^m(x)df(a-x)$$ and so your integral is: $$-\int_0^af^m(x)df(a-x)$$ if we let $u=(a-x)$ then $dx=-du$ and so: $$I=\int_a^0f^m(a-u)df(u)$$ $$=-\int_0^af^m(a-u)f'(u)du$$ as $a\to\infty$ I believe we have: $$I=-\int_0^\infty f_\infty^mf'(u)du=-f_\infty^m\left[f(u)\right]_0^\infty=f_\infty^mf(o)-f_\infty^{m+1}$$