Circular trap with integration by parts

Circular trap

When using integration by parts a second time, make sure you don't choose as the part to integrate the thing you got by differentiating the part to differentiate from the original product. Otherwise, you get in a circular trap and don't get any new information. The most typical application of integration by parts a second time is if you choose to differentiate again the expression that you already obtained through differentiation the first time.

Formal explanation of circular trap

Explanation using function notation

Consider an integration of the form:

$\int F(x)g(x)\,dx$

where $G$ is an antiderivative for $g$. Then:

$\int F(x)g(x)\,dx=F(x)G(x)-\int F'(x)G(x)\,dx$

Suppose that, for the new integral, we choose $G$ as the part to differentiate and $F'$ as the part to integrate. The expression then simplifies to:

$\int F(x)g(x)\,dx=F(x)G(x)-[G(x)F(x)-\int g(x)F(x)\,dx]$

Simplifying, we get:

$\int F(x)g(x)\,dx=\int F(x)g(x)\,dx$

In other words, we ended up with the original expression, and obtained no new information in the process. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.

Explanation using dependent-independent variable notation

Consider an integration of the form:

$\int u{\frac {dv}{dx}}\,dx$

Using integration by parts, write this as:

$\int u{\frac {dv}{dx}}\,dx=uv-\int v{\frac {du}{dx}}\,dx$

Now, let's say that, for the right side integral, we pick our "new" $u$ to be $v$ and our "new" $dv/dx$ to be $du/dx$. Then, our "new" $v$ is the "old" $u$ and we get:

$\int u{\frac {dv}{dx}}\,dx=uv-[vu-\int u{\frac {dv}{dx}}\,dx]$

Simplifying, we get:

$\int u{\frac {dv}{dx}}\,dx=\int u{\frac {dv}{dx}}\,dx$

In other words, we ended up with the original expression, and obtained no new information in the process. Another way of saying this is that we went in circles, i.e., went back along the path we came from, hence did not make any progress.