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When integrating by parts, at what point does the constant come in?

The rule has always been like this:

$$ \int u\,dv = uv - \int v\,du $$

The explanation is that this the "reversal" of the product rule:

$$ d(uv) = u\,dv + v\,du $$

But if that is the case, shouldn't a $C$ appear when you integrate both sides?

$$ uv = \int u\,dv + \int v\,du + C \quad ? $$

I've been told to just ignore the $C$ because it will appear in the final integral anyway, like this:

$$ \begin{align*} \int xe^x\,dx &= xe^x - \int e^x\, dx \\ &= xe^x - e^x + C \end{align*} $$

Fine. But what happens when integrating by parts results in the exact same integral as the original?

Like this, first no $C$:

$$ \begin{align*} \int e^x \sin x \, dx &= e^x \sin x - \int e^x \cos x \, dx \\ &= e^x \sin x - \left( e^x \cos x + \int e^x \sin x \, dx \right) \end{align*}$$

Then out of nowhere, a $C$ appears:

$$ \begin{align} 2\int e^x \sin x \, dx &= e^x \sin x - e^x \cos x + C \\ \int e^x \sin x \, dx &= \frac{1}{2} e^x \sin x - \frac{1}{2} e^x \cos x + C \end{align} $$

We know the final result has to contain a $C$, but where does it come from? No integral is being evaluated here, all we did was move the integral from the right to the left.

This is one thing that I think is not being properly taught.

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    $\begingroup$ In principle every time you integrate, you should add a constant, but if you ignore it and just add it in the end, the two final resulting expressions will be equivalent. $\endgroup$ – Git Gud Mar 16 '14 at 23:33
  • $\begingroup$ If you get a $C_1$ on the right and a $C_2$ on the left, you may combine them into one constant term. As such, it suffices to shove in a constant term at the very end. $\endgroup$ – Ian Coley Mar 16 '14 at 23:34
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    $\begingroup$ I agree with you that this thing is not properly taught. I am more radical, though, in that I think that it is the very symbol of "indefinite integral" that is flawed and confusing. It should be removed and kept only as a "guilt practice", like treating $dy/dx$ as a real fraction and things like that. $\endgroup$ – Giuseppe Negro Mar 16 '14 at 23:38
  • $\begingroup$ @GiuseppeNegro I'm very impressed that someone who works on differential equations thinks like this. Respect. $\endgroup$ – Git Gud Mar 16 '14 at 23:40
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    $\begingroup$ Possible duplicate of Constants of integration in integration by parts $\endgroup$ – Arnaud D. Sep 19 '18 at 9:26
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Here's one way to think about it: we reach this step: $$ \int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx $$ And for our next step, we'd like to add $\int e^x \sin x \, dx$ on both sides, so we do. $$ \int e^x \sin x \, dx + \int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx + \int e^x \sin x \, dx \\ \int [e^x \sin x + e^x \sin x] \, dx = e^x \sin x - e^x \cos x + \int [e^x \sin x - e^x \sin x] \, dx \\ \int 2[e^x \sin x] \, dx = e^x \sin x - e^x \cos x + \int 0\, dx \\ 2\int e^x \sin x \, dx = e^x \sin x - e^x \cos x + C $$ The intuition is that when we take the antiderivative twice in the same equation, we don't guarantee that the result is the same each time. So, the equation $$ \int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx $$ Is really something to the effect of $$ F(x) + C_1 = e^x \sin x - e^x \cos x - [F(x) + C_2] $$ Where $F$ is some antiderivative of $e^x \sin x$.

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  • $\begingroup$ Ah. That makes sense. $\endgroup$ – Dylan Mar 17 '14 at 0:17

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