Need help simplifying integral using integration by parts rule

Question

$$Y(\tau)=\int_{0}^{\tau} t\frac{dF}{dt}dt \:\:\:\:\:\: ; \:\:\: Y(0)=0$$ $$X(\tau)=F(\tau) \:\:\:\:\:\: ; \:\:\: X(0)=0$$

$$Q(X(\tau))=\int_{X(0)}^{X(\tau)} Y(t)dX$$

First real quick: Can I get away with $\int t \: dF=tF(t)$? (Pretty sure not)

Assuming the answer to this is no, I am trying to avoid an expression for $Q(X(t))$ that involves twice integrating, or any integration in general that will result in a huge complicated equation. For example, I prefer to deal with $\int F \: dt$ rather than $\int t \: dF$.

So, is it possible to use integration by parts to express $Q(X(\tau))$ in the following way:

$$Y(\tau)=\tau F(\tau)-\int_{0}^{\tau} F(t)dt$$

$$Q(X(\tau))=Y(\tau)X(\tau)-\int_{X(0)}^{X(\tau)} \frac{dY}{dX}dX$$

$$=Y(\tau)(X(\tau)-1)$$

Substituing in for $Y(\tau)$ and $X(\tau)=F(\tau)$:

$$Q(X(\tau))=(\tau F(\tau)-\int_{0}^{\tau} F(t)dt)(F(\tau)-1)$$

Is that allowed? I'm open to suggestions for better expressions.

Answer 1

No, this isn't right – too bad, you would have invented a generic method for solving all integrals :-) If

$$ \def\quest{\stackrel?=} \int_{X(0)}^{X(\tau)}Y(t)\,\mathrm dX\quest Y(\tau)(X(\tau)-1) $$

were correct, we could solve all integrals if we know a point $x_0$ where the integrand vanishes:

$$ \begin{align} \int_a^bf(x)\,\mathrm dx&=\int_{x_0}^bf(x)\,\mathrm dx-\int_{x_0}^af(x)\,\mathrm dx\\ &\quest f(b)(b-1)-f(a)(a-1)\;. \end{align} $$

The error is in the first step,

$$ \int_{X(0)}^{X(\tau)}Y(t)\,\mathrm dX\quest Y(\tau)X(\tau)-\int_{X(0)}^{X(\tau)}\frac{\mathrm dY}{\mathrm dX}\mathrm dX\;. $$

That's not a valid integration by parts. A valid integration by parts would look like this

$$ \int Y\mathrm dX=YX-\int X\mathrm dY $$

or like this if you prefer (with the same result):

$$ \int Y\mathrm dX=\int Y\frac{\mathrm dX}{\mathrm dt}\mathrm dt=YX-\int X\frac{\mathrm dY}{\mathrm dt}\mathrm dt=YX-\int X\mathrm dY\;. $$