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Suppose we have an expression of the follwoing form;

$\int_a^b fg $, and want to understand it. By integration by parts we have the following $\int_a^b fg = -\int_a^b Fg^{'} + Fg \mid$. But $\int_a^b Fg^{'}=\int_a^b \int f(t) dt \ g^{'}(x)dx$ which looks like a place to use Fubini. Are there any way to see how these theorem are related and if so are they related to the extent that one only has to keep one in mind? An example where one is favourable over the other would be good motivation.

Found this later ;

What is integration by parts, really?

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  • There is a similar question here: http://math.stackexchange.com/questions/85653/relation-between-integral-by-parts-and-fubinis-theorem – Alufat Nov 26 '15 at 22:54

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There is a connection but to make it clear you have to be more precise about the boundaries of the inner integral on the right hand side of your second equation. The primitive function $F$ of $f$ is only determined up to a constant. Assuming that it is chosen so that $F(a)=0,$ then we have

$$\int_a^bFg'=\int_{x=a}^b\int_{t=a}^xf(t)dt\ g'(x)dx$$

The integration takes place over a triangular area in the $(x,t)$ plane. If the hypotheses of Fubini's theorem are satisfied then we can look at the same triangle sideways and the double integral is equal to

$$\int_{t=a}^bf(t)\int_{x=t}^bg'(x)dx dt=\int_{t=a}^bf(t)[g(b)-g(t)]dt=F(b)g(b)-\int_a^b fg$$

which is the same as your integration by parts, taking into account that $F(a)g(a)=0$ by our choice of $F.$

If a different choice of $F$ is made, e.g., $F(0)=0,$ then a similar result can be obtained but the triangle becomes a trapezium and the intermediate expressions are slightly longer.

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