In Introduction to Linear Algebra: Fifth Edition by Gilbert Strang, it supposes that we have a matrix $A$ such that it differentiates the elements in a vector $x$ via $Ax$. It then resolves to find the transpose of $A$. First it defines the inner product as
$$x^{T}y = (x,y) = \int_{-\infty}^{\infty}x(t)y(t) dt \space\space\space (Eq \space 1)$$ Then: $$(Ax,y) = (Ax)^T y = x^T A^T y = x^T(A^T y) = (x,A^Ty) \space\space\space (Eq \space 2)$$ It then gives the following: 3) $$(Ax,y) = \int_{-\infty}^{\infty}\frac{dx}{dt}y(t) dt= \int_{-\infty}^{\infty}x(t)(-\frac{dy}{dt}) dt = (x,A^T)y \space\space\space (Eq \space 3)$$
And then it states that
I hope you recognize integration by parts. The derivative moves from the first function x(t) to the second function y(t). During that move, a minus sign appears. This tells us that the transpose of the derivative is minus the derivative
But I am confused by the "integration by parts". In order to do integration by parts for the second integral we would need: $$\int_{-\infty}^{\infty}\frac{dx}{dt}y(t) dt = x(t)y(t)|^\infty_{-\infty} - \int_{-\infty}^{\infty}x(t)(\frac{dy}{dt})dt = x(t)y(t)|^\infty_{-\infty} + \int_{-\infty}^{\infty}x(t)(-\frac{dy}{dt})dt$$
But to get $(Eq \space 3)$,$\space x(t)y(t)|^\infty_{-\infty}$ would need to be $0$. But how do we prove this? Am I missing something obvious?
Thanks.