Inquiry on an integration exercise. May you kindly assist me on
 the following simple-looking intergration exercise:
Let $f(x)$ be an analytic function of $x$ and $m$ be a real number.
Suppose that the integral of $f(x)x^m$ from $1$ to infinity (w.r.t $x$) exists and is equal to the integral of $f(x)x^{-m}$ over the same range. 
Can this equality be possible for any nonzero value of $m$ ?
My attempt : If $f(x)$ is of constant sign, then the answer is no: since for nonzero $m$, we will have a strict inequality between $f(x)x^m$ and $f(x)x^{-m}$, interpreting the integrals as area under the respective curves, it follows that the area under one graph is greater than the other.
Therefore, we remain with the case where $f(x)$ is an oscillating function, such as $f(x) = g(x)\sin(x)$.
 A: We are asking, can
$$\int_1^{\infty}f(x)x^mdx\stackrel{?}{=}\int_1^{\infty}f(x)x^{-m}dx$$
for $m\neq 0$. On the left hand side let $u=x^m$ and on the right hand side let $u=x^{-m}$ to get
$$\frac{1}{m}\int_1^{\infty}f(u^{\frac{1}{m}})u^{\frac{1}{m}}du\stackrel{?}{=}-\frac{1}{m}\int_1^{\infty}f(u^{-\frac{1}{m}})u^{-\frac{1}{m}}du$$
Then on the right hand side let $v=u^{-1}$, so that now we are considering if for $m\neq 0$ whether we can have
$$\int_1^{\infty}\frac{f(u^{\frac{1}{m}})u^{\frac{1}{m}}}{u^0}du\stackrel{?}{=}\int_1^{\infty}\frac{f(v^{\frac{1}{m}})v^{\frac{1}{m}}}{v^{2}}dv$$
Which equivalently put would mean we are able to find a function $g(x)$ such that
$$\int_1^{\infty}g (x)dx=\int_1^{\infty}\frac {g (x)}{x^2}dx \qquad  (1)$$
Then $f $ would be given by $f (x)=g (x^m)/x$. Notice that $(1) $ is satisfied iff
$$\int_1^{\infty}g (x)(1-x^{-2})dx=0\qquad (2)$$
Thus if we set $g$ to be of the form $g (x)=h'(x)/(1-x^{-2})$  where $h $ is a function such that $h (1)=\lim_{n\to\infty}h (n)$ then $(2) $ is satisfied and therefore so is $(1) $. Thus making functions $f $ that satisfy the property that we are interested in of the form
$$f (x)=\frac {h'(x^m)}{x (1-x^{-2m})}$$
