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I've encountered a problem by trying to use integration by parts to prove a certain theorem. I'm not sure if this is just confusion caused by notation, or I'm missing something important. The problem is proving the next line:

$ \int_a^x f'(t)dt = tf'(t)|_a^x - \int_a^xtf''(t)dt $

Let: $u=f'(t)$, and $dv=dt$. Therefore: $du=f''(t)$, and $v=t$.

Using integration by parts:

$\int_a^x f'(t)dt =\int_a^x udv= uv|_a^x -\int_a^xvdu=tf'(t)|_a^x-\int_a^xtf''(t)$

Notice that the last integral is missing the $dt$ symbol.

Where is the mistake in my logic? Is my understanding of notation wrong? Am I missing something obvious?

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    $\begingroup$ Rule of thumb: differentials always equal other differentials. So your $du$ should equal $f''(t)\,\color{red}{dt}$. $\endgroup$ – user137731 Oct 20 '16 at 2:58
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    $\begingroup$ Because $\dfrac{\mathrm d u}{\mathrm d t} = f''(t)$ ... $\endgroup$ – Graham Kemp Oct 20 '16 at 2:59
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If $$ u = f'(t) $$

then

$$ \mathrm{d}u = f''(t) \mathrm{d}t $$

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  • The integration by parts is quite easy to prove by the way. Let $F(x) = u(x)v(x)$, then $F'(x) = u'(x)v(x)+u(x)v'(x)$ and $u(b)v(b)-u(a)v(a) = F(b)-F(a) = \int_a^b F'(x)dx = \int_a^b u'(x)v(x)dx+\int_a^b u(x)v'(x)dx$. Therefore $$\int_a^b u(x)v'(x)dx = u(b)v(b)-u(a)v(a)- \int_a^b u'(x)v(x)dx$$

  • For the change of variable, it is as easy : let $g'(x) = f(x)$ and $\phi : [a,b] \to [c,d]$ inversible and $h(x) = g(\phi(x))$, then $h'(x)= \phi'(x) g'(\phi(x))= \phi'(x) f(\phi(x))$ and $$\int_a^b f(x)dx=g(b)-g(a) =g(\phi(\phi^{-1}(b)))-g(\phi(\phi^{-1}(a)) = h(\phi^{-1}(b))-h(\phi^{-1}(a))$$ $$ = \int_{\phi^{-1}(a)}^{\phi^{-1}(b)} h'(t)dt = \int_{\phi^{-1}(a)}^{\phi^{-1}(b)} f(\phi(t))\phi'(t)dt$$

Now the notations and the rules, such that $x = \phi(t)\implies dx = \phi'(t) dt$ are essentially made to be compatible with those properties of the Riemann integral.

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