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I was fooling around by differentiating some functions under the integral sign, and I seem to have stumbled on a problem I don't quite understand. There is another question here which asks something similar at the end, but the answer doesn't address it.

Consider the following integral: $$I = \int_{-\infty}^\infty \frac{\sin(k x)}{x} \text{d}x = \int_{-\infty}^\infty \frac{\sin(u)}{u} \text{d}u = \pi,$$

where I've used the substitution $u=kx$. Clearly, this integral is independent of $k$, and so $$\frac{\text{d}I}{\text{d}k} = 0.$$

However, if I use differentiation under the integral sign:

$$\frac{\text{d}I}{\text{d}k} = \int_{-\infty}^\infty\frac{\partial}{\partial k}\left( \frac{\sin(k x)}{x}\right) \text{d}x = \int_{-\infty}^\infty \cos(k x) \text{d}x,$$

where I have used Leibniz's Integral Rule, since the sinc function in the integral is continuous, and the integral converges. (Are there any assumptions that I'm missing here?) The problem is that this seems to imply that $$\int_{-\infty}^\infty \cos(k x) \text{d}x = 0,$$

which doesn't make sense! (Does it?) I study physics, so I have a habit of being slightly sloppy with mathematics. It wouldn't surprise me that there's some assumption that needs to be satisfied when differentiating under the integral sign that this function does not satisfy, but I can't figure out what it is. Can anyone tell me what I'm doing wrong?

EDIT:

Ok, so I have a slightly better idea of what I'm doing wrong, but I'm still not completely comfortable with it.

$$I = \int_{-\infty}^\infty \frac{\sin(k x)}{x} \text{d}x = \begin{cases} \,\,\pi& \quad k>0 \\-\pi& \quad k<0\end{cases} = \pi \,\, \text{sgn}(k),$$

where $\text{sgn}$ is the signum function. Using the fact that $$\frac{\text{d}}{\text{d}k} \text{sng}(k) = 2 \delta(k),$$ I get that

$$\frac{\text{d}I}{\text{d}k} = 2\pi \delta(k),$$ and so

$$\int_{-\infty}^\infty \cos(kx)\text{d} x = 2\pi \delta(k),$$

which seems to be the real part of the well known relation

$$\int_{-\infty}^\infty e^{i k x} \text{d}x = 2 \pi \delta(k).$$

So it looks like this might make sense if I think in terms of distributions. I'd appreciate any further input, though.

Philip
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  • $u$ is not independent of $k$.....so $I$ is also not. – Saket Gurjar Jun 16 '20 at 20:29
  • @SaketGurjar I really don't think that's true. As far as I know, the integral is $\pi$. Could you show me how it depends on $k$? – Philip Jun 16 '20 at 20:32
  • Consider : $f(x,y)=\sin(xy)$. Now if you use the substitution $u=xy$, $f(u)=\sin{u}$. Now think....is $f$ independent of $x$ due to this substitution?......$u$ still depends on $x$ – Saket Gurjar Jun 16 '20 at 20:34
  • The integral of the cosine over the real axis doesn't exist in the usual way, but why doesn't it make sense that if it is anything, then it may as well be $0$? – Allawonder Jun 16 '20 at 20:39
  • What I'm saying is that before the step where you wrote $\frac{dI}{dk}=0$, you claimed that "clearly this integral is independent of $k$".....this is what I feel is not true – Saket Gurjar Jun 16 '20 at 20:45
  • @SaketGurjar You're right, I just realised that it depends on the sign of $k$, but I still think that your argument was a little simplistic. – Philip Jun 16 '20 at 20:46
  • I agree with @Allawonder ... Look up the rule for differentiation under the integral sign. See what the hypotheses are. It cannot give you a divergent integral like $$\int_{-\infty}^\infty \cos(k x) \text{d}x$$ as the answer, can it? – GEdgar Jun 16 '20 at 20:46
  • @GEdgar That's exactly my question. I can't see anything wrong with what I've done, though I'm beginning to get an idea. – Philip Jun 16 '20 at 20:47
  • I don't think you have done anything wrong in differentiating the integral as you have shown....its just that i dont think its going in the direction of calculating the integral. – Saket Gurjar Jun 16 '20 at 20:49
  • @Allawonder I've realised something that I missed before, and it seems to be a little less problematic now. I've edited the question, but I'm still not completely happy with it. – Philip Jun 16 '20 at 20:59

1 Answers1

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Let's follow THIS explanation of integration under the integral sign.

Consider $$ I_{A,B}(k) := \int_A^B \frac{\sin{kx}}{x}\;dx, \qquad k>0 . $$ We will have to do limits $A \to -\infty$ and $B \to +\infty$ afterward.

We get $$ \frac{d}{dk}\;I_{A,B}(k) = \int_A^B \left[\frac{\partial}{\partial k} \frac{\sin{kx}}{x}{}\right]\;dx = \int_A^B\cos(kx)\;dx $$ BUT these limits all fail exist: $$ \lim_{A \to -\infty, B \to +\infty}\frac{d}{dk}\;I_{A,B}(k) \\ \lim_{A \to -\infty, B \to +\infty}\int_A^B \left[\frac{\partial}{\partial k} \frac{\sin{kx}}{x}\right]\;dx\\ \lim_{A \to -\infty, B \to +\infty} \int_A^B\cos(kx)\;dx $$ So the assertion that they are all equal is not much help.

Now, this limit does exist $$ \lim_{A \to -\infty, B \to +\infty} I_{A,B}(k) =\lim_{A \to -\infty, B \to +\infty}\int_A^B \frac{\sin{kx}}{x}\;dx $$ But (as we see here) you cannot use this phony rule $$ \frac{d}{dk}\lim_{A \to -\infty, B \to +\infty} I_{A,B}(k) =\lim_{A \to -\infty, B \to +\infty} \frac{d}{dk}I_{A,B}(k) $$ It can easily fail!

GEdgar
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  • Great! This actually answers my question as to what was wrong (I had a suspicion it had something to do with limits). But if you take the definition in the Wikipedia page you've linked, what exactly is it in the statement of the theorem that is not satisfied? – Philip Jun 16 '20 at 21:19
  • $f(x,k)$ is continuous, $f_x(x,k)$ is also continuous in both $x$ and $k$, and $A$ and $B$ being constants satisfy the other conditions. As you've shown, using the Leibniz rule here isn't valid, but does that mean that the linked theorem is incomplete? Do we also need that $f_t(x,t)$ be continuous, perhaps? – Philip Jun 16 '20 at 21:21
  • The theorem on that page is for $A$ and $B$ finite. Not $\pm \infty$. By writing this with $\pm\infty$ on there, you implicitly assumed (incorrectly) that the limit of the derivative is the derivative of the limit. – GEdgar Jun 16 '20 at 21:25
  • I see, it doesn't seem to be explicitly specified though, but I think I understand. Thanks! Just one further question if you have the time: could you take a look at my edit above? It seems like if I think of differentiation in the sense of distributions, I seem to get something that is plausibly sensible. What do you think? – Philip Jun 16 '20 at 21:31