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Is it possible to calculate this integral? I came across this integral while studying Fourier Analysis. And I can see that this function doesn't converge.

$$ \int_{-\infty}^{\infty} \cos (kt)\, dt $$

However, I came across the following equality and I wonder how to calculate it.

$$ \int_{-\infty}^{\infty} \cos (kt)\, dt = 2\pi \delta(k) $$

malachi
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  • depends on what for example? – malachi May 02 '25 at 12:14
  • We can't say for sure what the value will be at $\infty$. Even if we consider the Cauchy Principal value $(\lim_{n \to\infty} \int_{-n}^{n} \cos(kt))$, the limit doesn't exist – Random Math Enthusiast May 02 '25 at 12:20
  • Do you mean in the sense of a divergent renormalization? There probably is an answer then. Techniques in my answer here are applicable if you can turn the integral into a limit of a function (that doesn’t converge!) but it’s not obvious to me what the “natural” function should be whereas for sums there is a “natural” function. – Sidharth Ghoshal May 02 '25 at 12:49
  • I need to verify the algebra carefully but it seems your $2\pi \delta(k)$ formula is consistent with @anixx’s derivations here check the last 2 formula in that table specifically. – Sidharth Ghoshal May 02 '25 at 12:58

1 Answers1

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Short Answer: We can't

Long Answer: $\int_{-\infty}^{\infty}\cos(kt).dt$ as it is does not make sense. Plugging in infinity is not the same as plugging in a number

Perhaps if you said $\lim_{a\to -\infty, b\to\infty}\ \int_{a}^{b} \cos(kt).dt$ would be a little better.

But, $$I = \int_{a}^{b}\cos(kt).dt = \dfrac{\sin(kb)}{k} - \dfrac{\sin(ka)}{k}$$

$\lim_{a\to -\infty, b\to\infty} I$ does not exist

Even if we consider the Cauchy Principal Value,

$$\lim_{n \to \infty} \int_{-n}^{n} \cos(kt).dt = \lim_{n \to \infty}\dfrac{2}{k}\sin(kn)$$

This limit again does not exist

Update after your edit
The Dirac delta version, arises in the distributional sense used in Fourier analysis. In distribution theory, we define the integral by its action on test functions.

We have $$\int_{-\infty}^{\infty}e^{ikt}.dt = 2\pi\delta(k) $$

From there $$\int_{-\infty}^{\infty}\cos(kt).dt = \dfrac{1}{2} . 2\pi\delta(k) + \dfrac{1}{2} . 2\pi\delta(-k) = 2\pi\delta(k)$$

Note: The Dirac Delta function is an even function.

Conclusion
The integral is undefined classically but defined in distribution theory as a functional, not a number.

Random Math Enthusiast
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    This answer is confusing. You start with "this does not exist" and "we can't" (compute the integral), but then you suddenly give a value for it nevertheless. What is missing here is an explanation of what is going on and how we make sense of integrals like this. – Winther May 02 '25 at 13:40