Now we give an illustrative example where Cesàro works but the usual integral diverges.

Take \[ f(x)=\cos x. \] Usual improper integral diverges. As T → ∞, 2 sinT oscillates and has no limit, so \[ \int _{-\infty }^{\infty }\cos x\, dx \] does not exist in the usual improper sense.

However, its Cesàro integral converges. Indeed, compute the Cesàro mean \[ C(T) = \frac{1}{T}\int_0^T S(t)\, {\text d}t = \frac{2 \left( 1 - \cos T\right)}{T}, \qquad S(t) = \int_{-t}^t \cos x\,{\text d} x = 2\,\sin t . \]

Usual improper integral over ℝ diverges.
Cesàro integral over ℝ exists and equals 0.

Another Example where both notions agree.

Take \[ f(x) = e^{-x^2}. \] The usual improper integral is classical: \[ \int _{-\infty }^{\infty }e^{-x^2}\, {\text d} x = \sqrt{\pi }. \]

• Usual improper integral: \( \displaystyle \quad \int _{-\infty }^{\infty } e^{-x^2} \,{\text d}x = \sqrt{\pi }. \)
• Cesàro integral: exists and equals the same value √π.

A Cesàro-type inversion formula says that for f ∈ 𝔏¹(ℝ). \[ f(x) = \frac{1}{2\pi }\lim _{T\rightarrow \infty }\,\frac{1}{T}\int_0^T {\text d}t \int _{-t}^te^{{\bf j}x\xi }\, \hat {f}(\xi )\, {\text d}\xi \] for almost every x, including all points of continuity of f. So instead of a single symmetric truncation \( \displaystyle \quad \int _{-T}^T, \quad \) we take the Cesàro mean of these truncations.

Rewriting Cesàro inversion as convolution with an approximate identity. Start from \[ \frac{1}{T}\int _0^T {\text d}t\, \int _{-t}^t \,e^{{\bf j}x\xi }\, \hat {f}(\xi )\, {\text d}\xi . \] Insert the definition of Fourier transform ℱ[f]: \[ \hat {f}(\xi ) =\int_{\mathbb{R}} e^{-{\bf j}y\xi }\,f(y)\, {\text d}y. \] Then \begin{aligned}\frac{1}{T}\int _0^T {\text d}t\,\int _{-t}^t \,e^{{\bf j}x\xi }\, \hat {f}(\xi )\, {\text d}\xi &= \frac{1}{T}\int _0^T {\text d}t \,\int_{-t}^t \,e^{{\bf j}x\xi }\left( \int _{\mathbb{R}}e^{-{\bf j}y\xi } \,f(y)\, {\text d}y\right) {\text d}\xi \\ &=\frac{1}{T}\int _0^T{\text d}t \,\int_{\mathbb{R}} \,f(y)\left( \int _{-t}^t \,e^{{\bf j}(x-y)\xi }\, {\text d}\xi \right) {\text d}y\, .\end{aligned} Compute the inner integral: \[ ???? \] Thus, \[ \frac{1}{T}\int _0^T\int _{-t}^te^{ix\xi }\, \hat {f}(\xi )\, {\text d}\xi \, {\text d}t = \int _{\mathbb{R}}f(y)\, K_T(x-y)\, {\text d}y, \] where the Cesàro kernel is \[ K_T(u):=\frac{1}{T}\int _0^T\frac{2\sin (tu)}{u}\, dt=\frac{2}{u}\cdot \frac{1}{T}\int _0^T\sin (tu)\, dt. \] Compute the last integral: \[ \int _0^T\sin (tu)\, {\text d}t = \frac{1-\cos (Tu)}{u}, \] so \[ ????? \] Therefore, the Cesàro inversion formula can be written as \[ f(x)=\frac{1}{2\pi }\lim _{T\rightarrow \infty }(f*K_T)(x) \] for almost every x, where (K_T) (T>0) is an approximate identity.

Explicit example: indicator of an interval. We take \[ f(x) =\mathbf{1}_{[-1,1]}(x) =\left\{ \, \begin{array}{ll}\textstyle 1,&\textstyle |x|\leq 1,\\ \textstyle 0,&\textstyle |x|>1.\end{array}\right. \] This function is integrable, has jump discontinuities at ± 1, and is continuous elsewhere. Fourier transform of f is \[ \hat {f}(\xi ) =\int _{-1}^1 \,e^{-{\bf j}x\xi }\, {\text d}x = \left[ \frac{e^{-ix\xi }}{-{\bf j}\xi }\right]_{x=-1}^{x=1} =\frac{e^{-{\bf j}\xi }-e^{{\bf j}\xi }}{-{\bf }\xi } = 2\, \frac{\sin \xi }{\xi }. \] So the formal inverse transform is \[ ????? \] This integral is only conditionally convergent and must be interpreted carefully (e.g., as a principal value or via summability methods).

Cesàro inversion for this function. Apply the Cesàro inversion formula: \[ f(x)=\frac{1}{2\pi }\lim _{T\rightarrow \infty }\frac{1}{T}\int_0^T\int_{-t}^t \,e^{{\bf j}x\xi }\, \hat {f}(\xi )\, {\text d}\xi \, {\text d}t. \] Using the convolution form, \[ \frac{1}{2\pi }\frac{1}{T}\int _0^T\int _{-t}^te^{ix\xi }\, \hat {f}(\xi )\, d\xi \, {\text d}t = \frac{1}{2\pi }\int_{\mathbb{R}}f(y)\, K_T(x-y)\, {\text d}y = \frac{1}{2\pi }\int _{-1}^1 \,K_T(x-y)\, {\text d}y. \] So the Cesàro approximants are \[ f_T(x) := \frac{1}{2\pi }\int _{-1}^1 \,K_T(x-y)\, {\text d}y = \frac{1}{2\pi }\int_{x-1}^{x+1} \,K_T(u)\, {\text d}u. \] Now note the key properties of K_T:

Normalization: \[ \int _{\mathbb{R}}K_T(u)\, {\text d}u = 2\pi \quad \mathrm{for\ all\ }T>0. \] (So K_T/(2π) has total mass 1.)

Approximate identity: as T → ∞, \[ \frac{1}{2\pi }K_T(u)\, du\rightharpoonup \delta _0\quad \mathrm{in\ the\ sense\ of\ distributions,} \] and more explicitly, K_T/(2π) concentrates near u=0 and its mass outside any fixed neighborhood of 0 tends to 0. Therefore, for any point of continuity x of f, \[ \lim _{T\rightarrow \infty }f_T(x)=\lim _{T\rightarrow \infty }\frac{1}{2\pi }\int _{-1}^1K_T(x-y)\, dy=f(x). \] In our example: If |x|<1, then f is continuous at x with value 1, and \[ \lim _{T\rightarrow \infty }f_T(x)=1. \] If |x|>1, then f is continuous at x with value 0, and \[ \lim _{T\rightarrow \infty }f_T(x)=0. \] At the jump points x = ± 1, one can show (by symmetry of the kernel) that exactly as in the classical Fourier series/Fejér kernel situation. So the Cesàro inversion recovers f almost everywhere and gives the expected “midpoint of the jump” at discontinuities.

Comparison with ordinary (non-Cesàro) inversion. If instead you use the plain symmetric truncation \[ f_R(x) :=\frac{1}{2\pi }\int _{-R}^Re^{{\bf j}x\xi }\, \hat {f}(\xi )\, {\text d}\xi , \] you obtain:
If |x|<1, then f is continuous at x with value 1, and \lim _{T\rightarrow \infty }f_T(x)=1. If |x|>1, then f is continuous at x with value 0, and \( \displaystyle \quad \lim _{T\rightarrow \infty }f_T(x)=0. \)

At the jump points x = ± 1, one can show (by symmetry of the kernel) that f(1) = ½, exactly as in the classical Fourier series/Fejér kernel situation. So the Cesàro inversion recovers f almost everywhere and gives the expected “midpoint of the jump” at discontinuities.

Comparison with ordinary (non-Cesàro) inversion. If instead you use the plain symmetric truncation \[ f_R(x) := \frac{1}{2\pi }\int_{-R}^R \,e^{{\bf j}x\xi }\, \hat {f}(\xi )\, {\text d}\xi , \] you obtain ======================================== Conclusion (Cesàro Inversion via Characteristic Function) For the inverse Fourier transform, the Cesàro (C,1) method replaces the sharp cutoff \chi _{[-R,R]}(\xi ) with the Fejér–type averaging \frac{1}{R}\int _0^R\chi _{[-r,r]}(\xi )\, dr=\left( 1-\frac{|\xi |}{R}\right) _+. This produces the tempered Cesàro kernel K_R(x)=\frac{1}{2\pi }\int _{-R}^R\left( 1-\frac{|\xi |}{R}\right) e^{ix\xi }\, d\xi =\frac{1}{\pi }\frac{1-\cos (Rx)}{x^2}. The key point is: K_R is positive, integrable, and forms an approximate identity. Therefore, for every f\in L^1(\mathbb{R}) (or tempered distribution), f(x)=\lim _{R\rightarrow \infty }(f*K_R)(x) at every Lebesgue point of f. This is the Cesàro inversion theorem for the Fourier transform. Conclusion (Cesàro Inversion via Characteristic Function) For the inverse Fourier transform, the Cesàro (C,1) method replaces the sharp cutoff \chi _{[-R,R]}(\xi ) with the Fejér–type averaging \frac{1}{R}\int _0^R\chi _{[-r,r]}(\xi )\, dr=\left( 1-\frac{|\xi |}{R}\right) _+. This produces the tempered Cesàro kernel K_R(x)=\frac{1}{2\pi }\int _{-R}^R\left( 1-\frac{|\xi |}{R}\right) e^{ix\xi }\, d\xi =\frac{1}{\pi }\frac{1-\cos (Rx)}{x^2}. The key point is: K_R is positive, integrable, and forms an approximate identity. Therefore, for every f\in L^1(\mathbb{R}) (or tempered distribution), f(x)=\lim _{R\rightarrow \infty }(f*K_R)(x) at every Lebesgue point of f. This is the Cesàro inversion theorem for the Fourier transform.    ■

1. Cossar, J., The Cesàro Summability of Fourier Integrals, Proceedings of the Edinburgh Mathematical Society, Vol. 7, Issue 2 (1945), pp. 84–92. doi: 10.1017/S0013091500024366