Check that the operators are inverse to each other Let $f$ be a differentiate function and consider two operators
$$
A(f(x))=\int_0^1 \frac{ d}{dx}f(x t^\mu) dt,\\
B(f(x))=\mu xf(x)+\int_0^x f(t) dt,
$$
where $\mu $ is a parameter.
I need to prove that
$$
A \circ  B =B \circ A = I
$$
where $I$ is  the identity operator $I(f(x))=f(x).$
It is easy to prove it  for powers $x^n$ and then  for power series but it would be interesting to perform direct calculations and prove  it for any  function $f.$
But I was confused with calculation of $A(B(f(x))$.
Any help?
 A: I believe it is necessary that $f(0)=0$. For example, if $f(x)=1$ for all $x$, then $B\circ A(f(x))=0$. Indeed this is true for $f(x)=C$ for any constant $C$. Therefore, $A$ is not injective. Moreover, for any $f$, the function $g(x)=B(f(x))$ satisfies $g(0)=0$, so $B$ is not surjective. Both these issues are resolved if the space of differentiable functions we consider are those such that $f(0)=0$. Additionally, the proof below assumes $\mu\neq -1$.
\begin{align}
    A\circ B(f(x))&=\int_0^1\frac{d}{dx}\left(\mu x s^\mu f(xs^\mu)+\int_0^{xs^\mu} f(t)\,dt\right)\,ds\\
&=\int_0^1 \mu s^\mu f(xs^\mu)+\mu x s^{2\mu} f'(xs^\mu)+s^\mu f(xs^\mu)\,ds.
\end{align}
Let $y=xs^\mu$. Then $ds=\frac{1}{\mu}\frac{y^{1/\mu-1}}{x^{1/\mu}}dy$ and $s=(y/x)^{1/\mu}$. With these substitutions we have
\begin{equation}
A\circ B(f(x))=\int_0^x \left(1+\frac{1}{\mu}\right)\frac{y^{1/\mu}}{x^{1+1/\mu}}f(y)+\frac{y^{1+1/\mu}}{x^{1+1/\mu}}f'(y)\,dy.
\end{equation}
Let $\gamma=1/\mu$. Then
\begin{align}
A\circ B(f(x))&=\frac{1}{x^{\gamma+1}}\int_0^x(1+\gamma)y^\gamma f(y)+y^{\gamma+1}f'(y)\,dy\\
&=\frac{1}{x^{\gamma+1}}\int_0^x\frac{d}{dy}\left(y^{\gamma+1} f(y)\right)\,dy\\
&=\frac{1}{x^{\gamma+1}}\left(x^{\gamma+1}f(x)-0\right)\\
&=f(x).
\end{align}
Now
\begin{align}
B\circ A(f(x))&=\mu x \int_0^1 \frac{d}{dx}f(xt^\mu)\,dt+\int_0^x \int_0^1\frac{d}{ds}f(st^\mu)\,dt\,ds\\
&=\mu x \int_0^1 \frac{d}{dx}f(xt^\mu)\,dt+\int_0^1 \int_0^x\frac{d}{ds}f(st^\mu)\,ds\,dt\\
&=\mu x \int_0^1 t^\mu f'(xt^\mu)\,dt+\int_0^1 f(xt^\mu)-f(0)\,dt\\
&=\int_0^1 \mu xt^\mu f'(xt^\mu)+f(xt^\mu)\,dt.
\end{align}
Note that in the last line we use $f(0)=0$. Let $y=xt^\mu$. Then $dt=\frac{1}{\mu}\frac{y^{1/\mu-1}}{x^{1/\mu}}dy$ and $t=(y/x)^{1/\mu}$. Then
\begin{align}
B\circ A(f(x))&=\int_0^x\frac{y^{1/\mu}}{x^{1/\mu}}f'(y)+\frac{1}{\mu}\frac{y^{1/\mu-1}}{x^{1/\mu}}f(y)\,dy\\
&=\frac{1}{x^\gamma}\int_0^x y^\gamma f'(y)+\gamma y^{\gamma-1} f(y)\,dy\\
&=\frac{1}{x^\gamma}\int_0^x \frac{d}{dy}\left(y^\gamma f(y)\right)\,dy\\
&=\frac{1}{x^\gamma}\left(x^\gamma f(x)-0\right)\\
&=f(x).
\end{align}
A: We have
$B(f)(xt^\mu) = \mu x t^\mu f(xt^\mu) + \int_0^{xt^{\mu}}f(s)ds$
$\dfrac{d}{dx}B(f)(xt^\mu) = \mu t^\mu f(xt^\mu) + \mu x t^{2\mu}f^\prime(xt^\mu) + t^\mu f(xt^\mu) = (\mu + 1)t^\mu f(xt^\mu) + \mu x t^{2\mu}f^\prime(xt^\mu) = \dfrac{d}{dt}(t^{\mu + 1}f(xt^\mu))$
So
$(A \circ B)(f)(x) = \displaystyle\int_0^1 \dfrac{d}{dt}(t^{\mu + 1}f(xt^\mu))dt = \left[t^{\mu + 1}f(xt^\mu)\right]_{t = 0}^1 = f(x).$
