$$\int_0^l{\dfrac{2\sin^3\left(\frac{{\pi}x}{l}\right)\sin\left(\frac{{\pi}nx}{l}\right)}{l}dx} = \dfrac{12\sin\left({\pi}n\right)}{{\pi}\left(n^4-10n^2+9\right)}$$

But this function is not defined at $n = 1$ and $n = 3$. Why does this happen even though I integrated it for all $n$. This also clearly has an integral defined if we substitute $n = 1$ before integrating.

link to the steps while integrating. Its an online integral calculator

  • $\begingroup$ Are you missing the variable of integration? $\endgroup$ – Andrew Li Apr 23 '18 at 15:24
  • $\begingroup$ Show us your development and we will tell you where the "for all $n$" claim is invalidated. $\endgroup$ – Yves Daoust Apr 23 '18 at 15:27
  • $\begingroup$ What did you mean by definition? Is it the steps I used to reach at this answer? $\endgroup$ – Allen Apr 23 '18 at 15:30
  • $\begingroup$ @YvesDaoust I have added the link for calculation steps $\endgroup$ – Allen Apr 23 '18 at 15:33
  • $\begingroup$ if you take a look at the steps, you'll see that there are several substitutions that won't work for particular values of $n$ (i.e.$v=(n+3)u$ does not work for $n=-3$). That's why you can't use this formula for $n=-3, -1, 1, 3$. $\endgroup$ – Vasya Apr 23 '18 at 15:37

You are probably trapped by a case similar to

$$I_m=-\int_0^{2\pi}\cos mx\,dx=\left.\frac{\sin mx}m\right|_0^{2\pi}=\frac{\sin2\pi m}m,$$

believing that it is established "for all $m$".

The final expression is indeed undefined for $m=0$ because the integrand degenerates to $1$ and the antiderivative doesn't hold.

  • $\begingroup$ Why does the antiderivative not hold. Is it because we are wrongly picking the antiderivative $\endgroup$ – Allen Apr 23 '18 at 15:43
  • $\begingroup$ @Allen: this is obvious. Set $m=0$. $\endgroup$ – Yves Daoust Apr 23 '18 at 15:44
  • $\begingroup$ Yes it is obvious that the final expression is not defined. But why does this happen, shouldn't the antiderivatiive define it at all points $\endgroup$ – Allen Apr 23 '18 at 15:46
  • $\begingroup$ @Allen Not for all values of parameters. Consider the antiderivative $\frac{x^{n+1}}{n+1}$ of $x^n$. Why does not it work for $n=-1$? $\endgroup$ – user Apr 23 '18 at 15:56
  • $\begingroup$ @Allen: by virtue of what principle should it ? $\endgroup$ – Yves Daoust Apr 23 '18 at 16:01

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