The following problem is from The Calculus of Variations by B.von Brunt (page 215, Exercise 9.2.1)
Let $$ J(y)=\int_a^b xy'^2\mathrm{d}x. $$
Show that the transformation $$ X=x+\epsilon2x\ \mathrm{ln}x,\ \ Y=(1+\epsilon)y $$ is a variational symmetry for $J$.
My Attempt: I know to be a variational symmetry we require; $$ J(Y)=\int_{X(a)}^{Y(b)}X\dot{Y}^2\mathrm{d}X=J(y)=\int_a^b xy'^2\mathrm{d}x. $$ where $\dot{Y}=\displaystyle\frac{\mathrm{d}Y}{\mathrm{d}X}$
ie. The transformation does nothing to the value of the integral.
It has been proven that both $X$ and $Y$ have inverse functions. Hence, \begin{align} Y(X)&=(1+\epsilon)y(x(X))\\ \dot{Y}(X)&=(1+\epsilon)\displaystyle\frac{dy}{dx}\frac{dx}{dX} \end{align} also, \begin{align} \displaystyle\frac{dx}{dX}&=\frac{1}{1+2\epsilon(\ln{x}+1)} \end{align} but when I plug all this in I do not get the required equivalence. What am I doing wrong?
Any help/guidance would be great. Thanks.
This is what I have
$$ \frac{dY}{dX} = \frac{dY/dx}{dX/dx} = \frac{d((1+\epsilon)y)/dx}{d(x+\epsilon 2x \ln(x))/dx} = \frac{(1+\epsilon)\dot{y}}{1+2\epsilon + 2\epsilon\ln(x)} $$
so
$$ \int^{Y(b)}_{X(a)} X \dot{Y}^2 dX = \int_a^b (x+\epsilon 2x \ln(x)) \left(\frac{(1+\epsilon)\dot{y}}{1+2\epsilon + 2\epsilon\ln(x)}\right)^2 \frac{dX}{dx}dx $$
$$ = \int_a^b x \frac{(1+\epsilon 2 \ln(x))(1+\epsilon)^2}{1+2\epsilon + 2\epsilon\ln(x)}\dot{y}^2dx $$
Now since the term $$ \frac{(1+\epsilon 2 \ln(x))(1+\epsilon)^2}{1+2\epsilon + 2\epsilon\ln(x)} = \frac{(1+\epsilon 2 \ln(x))(1+2\epsilon+\epsilon^2)}{1+2\epsilon + 2\epsilon\ln(x)} $$ $$ = \frac{1+2\epsilon + 2\epsilon\ln(x)}{1+2\epsilon + 2\epsilon\ln(x)} + \mathcal{O}(\epsilon^2) $$
Then looking at the Theorem as stated in this book it follows that the functional is variational invariant under the transformation.