For recreational purpose, I am searching for functions that satisfy $$\left(\int_a^t f(x) dx\right)^k= \int_a^t f(x)^k dx$$ for some $k\ne 0, 1$, $a$ is a constant and $t$ is a variable(in other words, the integrals are indefinite).
I have just found out that if $f(x)$ is a monomial $Ax^n$, the only possible solution is $n=0$ which makes it trivial.
Can anyone come up with some non trivial $f(x)$s?
On
Let $ \int_{a}^{t} f(x) dx = F_a(t)$
Then $ ((F_a(t))^k)' = k F_a(t)^{k-1}f(x) $
Moreover $ (\int_{a}^{t} f(x)^k dx)' = f(t)^k $
So we have that $ k F_a(t)^{k-1}f(x) = f(t)^k $
$$ k F_{a} (t)^{k-1} = f(t)^{k-1}$$
Swapping out $f(t)$ with $F_a(t)'$ and letting $F_a(t) = u(t) $ (dropping the t for conveniene). This is
$$ ku^{k-1} = (u')^{k-1}$$
$$ k^{\frac{1}{k-1}} u = u'$$
Yielding $u(t) = Ce^{k^{\frac{1}{k-1}t}} $
As a general solution.
As is often the case with integral/differential equations, exponentialfunctions are a good place to start. Suppose $a=-\infty$ (to eliminate some problematic terms), and $f(x)=e^{rx}$. $$ \left(\int_a^tf(x)dx\right)^k=\left(\int_{-\infty}^te^{rx}dx\right)^k=\frac{1}{r^k}e^{rkt} $$ $$ \int_a^t\left(f(x)dx\right)^k=\int_{-\infty}^te^{rkx}dx=\frac{1}{rk}e^{rkt} $$ We see that these two will be the same provided $rk=r^k$, or equivalently $r=k^{1/(k-1)}$. For the specific case of $a=-\infty$, we have $f(x)=Ae^{k^{1/(k-1)}x}$.