Analysis question on Integration bounds I have to find all functions f(t) such that
$\int_x^{x^2} f(t) d t=\int_1^x f(t) dt$
I think the solutions are all the functions of the form a/(x+b) because the logarithm would divide the two powers and ln(1) is 0 so that would go away..
Not sure if there are any others though..
 A: We are going to assume that the function is well defined on the interval $(1,x)$ and $(x,x^2)$ since it is integrable there apiori.  
$$\int_x^{x^2} f dt = \int_x^1 f dt + \int_1^{x^2} f dt  = \int_1^x f dt$$
The only thing that we need to worry about is if the integral is well defined on $(1, x^2)$. If $x>1$ then it is the case, because $(1,x^2)$ is the disjoint union of the other two well defined integrals.  If $x=0$, we only get the trivial result (so we always ignore this).  If $x<0$ then $(x^2,1) \subset (x,1)$, so it is still well-defined.  This yields the following result
$$\int_1^{x^2} f dt = 2 \int_1^x f dt $$
From here, I recommend you look up the definition of the weak derivative, this should yield a nice answer (if you want to assume for example that your function is not even continuous)
A: We must make the assumption that f(t) is defined at least on the interval (1, $x^2$) and integrable there.  Then $\int_x^{x^2}f(t)dt = \int_1^{x^2}f(t)dt - \int_1^xf(t)dt$.   
By hypothesis  $\int_x^{x^2}f(t)dt =  \int_1^xf(t)dt$.   So we have
 $\int_1^{x^2}f(t)dt = 2\int_1^xf(t)dt \hspace{50px}$ (1)
Assuming f is continuous we can differentiate both sides of (1) to get f($x^2$) = 2f(x).  As per TimG's thought the answer here is f(x) = log(x). $\hspace{100px}$ (2)
This log can be to any base, so there is a family of functions saisfying the condition.
This leaves two questions:  what to do if f is not continuous; and even if f is continous are there any other solutions to (2). 
There are two kinds of "not continuous" -- discontinous at a finite or countably many number of  discrete points; or discontinuous at uncountably many points.  For the first case, we can still go with the solution f(x) = log(x) (piecewise), because those discontinuities do not affect the integral.
In the event of uncountably many discontinuities, the function is not Riemann integrable.  It may be Lebesgue integrable, and if anyone wants to address this question they are welcome, but I don't know.
Now about other solutions.  If there is a g(x) which is another solution it must satisfy $\int_1^x g(t)dt = \int_1^x log(t)dt$.  Now if g is continuous we can differentiate both sides and conclude g(x) = log x.  But we can construct a discontinuous g (other than a log x with holes) which satisfies the condition.  Picking a log to a specific base we set
$\begin{pmatrix}
g(x) = logx -\epsilon & t < (x-1)/2\\
g(x) = logx +\epsilon& t \ge (x-1)/2\\
\end{pmatrix}$
The integral of g will be the same as that of f. This opens up endless possibilities.
A: Assuming that the equality is true $\forall x \in \mathbb{R}$, differetiating both sides by $x$ and applying the Fundamental Theorem of Calculus:
$$\frac{d}{dx}\left(\int_x^{x^2} f(t) d t\right)=\frac{d}{dx}\left(\int_1^x f(t) dt\right)$$
$$\frac{d}{dx}(x^2)f(x^2)-\frac{d}{dx}(x)f(x)=\frac{d}{dx}(x)f(x)$$
$$2xf(x^2)-f(x)=f(x)$$
$$xf(x^2)=f(x)$$
So we obtain the equation $xf(x^2)=f(x)$.
