Sometimes when doing integration by parts I noticed that some book authors write this kind of derivation.

$$\int f(x)\ dx = some\ computations\ here = G(x) - \int f(x)\ dx \tag{1}$$

And now... from here they conclude that

$$2 \cdot \int f(x)\ dx = G(x) \tag{2} $$

so it follows that

$$\int f(x)\ dx = \frac{G(x)}{2} \tag{3}$$

But I feel something is not quite rigorous here. First of all what does it mean $2$ times an indefinite integral (that's the expression we have in $(2)$?!). This expression is not defined, right? Is it OK to write it then, or to conclude that $(3)$ follows.

Also... in this whole derivation from $(1)$ to $(2)$ to $(3)$, what happens with the free constant $ H(x) + C$ which we usually write when solving indefinite integrals?

These two items seem to confuse me. I understand the answer at the end comes out correct, but I feel like the derivation is somewhat unjustified (not rigorous). Could someone clarify this?

  • $\begingroup$ A definite integral is a function. What is the problem multiplying a function by $2$ ? $\endgroup$ – Yves Daoust Aug 28 '20 at 12:17
  • $\begingroup$ @YvesDaoust Indefinite integral. It is not really a function, is it? I think it's any primitive function. So... it's like a family of functions, no? $\endgroup$ – peter.petrov Aug 28 '20 at 12:18
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    $\begingroup$ What is the problem multiplying a family of functions by $2$ ? :-) $\endgroup$ – Yves Daoust Aug 28 '20 at 12:19
  • $\begingroup$ @YvesDaoust Well :) That's what is confusing me, I guess. How do we do algebraic manipulations just like that without clearly stating what we mean?! Seems like something is implicitly covered up here. $\endgroup$ – peter.petrov Aug 28 '20 at 12:19
  • $\begingroup$ @peter.petrov If you have any doubt about indefinite integrals, add bounds to them ! $\endgroup$ – TheSilverDoe Aug 28 '20 at 12:21

Your concern should vanish if you consider

$$\int f(x)\,dx:=\int_{x_0}^x f(t)\,dt$$ for some arbitrary $x_0$.

  • $\begingroup$ Interesting... Hm... But aren't we chasing our tails then (i.e. going in a loop)? To define the expression on the right side and to prove its derivative is $f(x)$, aren't we using some indefinite integrals apparatus? If so then this definition would not work, right? $\endgroup$ – peter.petrov Aug 28 '20 at 12:24
  • $\begingroup$ @peter.petrov: redo your initial computation with this formulation. You will see all problems vanish. $\endgroup$ – Yves Daoust Aug 28 '20 at 12:30

For the incredulous, here is a bare example, without constants:

$$\int x^2\,dx=\int x\cdot x\,dx=\frac{x^2}2x-\int \frac{x^2}2dx$$ so that

$$\frac32\int x^2\,dx=\frac{x^3}2.$$

But you may add constants wherever you want,

$$\int x\cdot x\,dx+C_1=\frac{x^2}2x+C_2-\left(\int \frac{x^2}2dx+C_3\right)$$ you still get

$$\int x^2\,dx=\frac{x^3}3+C.$$

  • $\begingroup$ Thanks... this somewhat clears my doubts. I think I get it better now. $\endgroup$ – peter.petrov Aug 28 '20 at 13:18

As I understand it, these expressions are a "general understanding" of an application for integration by parts. I've had this same doubt in the past. The idea is that you use the integration by parts twice and then get the desired integral, without having to calculate it explicitly.

I will give you an example here where this occurs. This example is very useful!

Consider $f(x)=e^{x}\cos x$:

$$ \int e^{x}\cos x\, dx = e^{x}\sin x-\int e^{x}\sin x\, dx = e^{x}\sin x+e^x\cos x-\int e^{x}\cos x \, dx $$

Note that the desired integral appears on both sides of the above equality. So we have:

$$ \int e^{x}\cos x\, dx + \int e^{x}\cos x\, dx = e^{x}\sin x+e^x\cos x $$


$$ 2\int e^{x}\cos x\, dx = e^{x}\sin x+e^x\cos x $$


$$ \int e^{x}\cos x\, dx = \frac{e^{x}\sin x+e^x\cos x}{2} $$

Here your $G(x) = e^{x}\sin x+e^x\cos x$


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