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For $f(x)=\cos x$ on $[-a,a]$ and zero outside, the angular-frequency Fourier transform follows by writing the cosine as two complex exponentials. Combining fractions gives $[2k\cos a\sin (ak)-2\sin a\cos (ak)]/(k^2-1)$. The apparent singularities at $k=\pm 1$ are removable, with value $a+\sin (2a)/2$. If $\cos a\ne 0$, jumps at the endpoints give an oscillatory $O(|k{|}^{-1})$ envelope. If $\cos a=0$, the function is continuous at the endpoints but its first derivative jumps, giving an $O(|k{|}^{-2})$ envelope. This is an explicit instance of Fourier decay from a derivative jump.

Use the standard Fourier transform on the whole real axis. The printed lower limit is missing a minus sign. For this integrable even function, its only obstruction to smoothness is the origin, so its high-frequency algebraic terms come from its local cusp:Localize near zero with a smooth cutoff. The smooth polynomial terms, including $|x{|}^6=x^6$, contribute no algebraic cusp term; their localized transforms decay faster than any prescribed inverse power. The supplied half-line formula, interpreted away from $k=0$ in an Abel-regularized or tempered distribution sense, gives the Fourier transform of an algebraic cuspFor $p=3$ the cosine is $1$, and for $p=9$ it is $-1$. Both cusp coefficients in the local expansion are $-1$. HenceThe second coefficient is $29!=725760$. The absence of a $|k{|}^{-7}$ term follows from the smooth even power $x^6$, not from neglecting an available correction.

One can verify the signs by integration by parts on $2\mathrm{Re}{\int }_0^{\infty }{e}^{-ikx}(1+x^3{)}^{-1}dx$. Its endpoint derivatives first have nonzero relevant values $f^{(3)}(0)=-3!$ and $f^{(9)}(0)=-9!$, which contribute twice those values divided by $(ik{)}^4$ and $(ik{)}^{10}$. Higher derivatives are integrable on the half-line, justifying the displayed remainder after further integrations. Equivalently these coefficients are the Fourier decay from a derivative jump at the origin.

The input is even, so its Fourier transform is real and even. Decomposing the cosine into complex exponentials and integrating over its support givesCombining the fractions and using the angle addition formula yieldsThe apparent poles at $k=\pm 1$ are removable singularities; either the integral or the sine-quotient limit gives $\tilde{f}(\pm 1)=a+\sin (2a)/2$.

If $\cos a\ne 0$, expansion of the denominator givesThe oscillatory envelope is therefore $O(|k{|}^{-1})$. At $a=\pi /2+n\pi$, $n=0,1,2,\dots$, the endpoint values vanish and the first term cancels:This gives an $O(|k{|}^{-2})$ envelope. At general $a$ the truncation creates jumps in the function; at these special values it leaves the function continuous but creates jumps in its first derivative. Each additional degree of endpoint smoothness gives another inverse power, as explained by Fourier decay from a derivative jump. The exact example is the Fourier transform of a truncated cosine; isolated zeros of its oscillatory factors do not change the envelope order.