Extend $f$ to be zero for $x<0$. Define $h(x)=0$ for $x>0$ andThen the full-line equation is $K\ast f=f+h$. Use the Fourier transform conventionUnder the corresponding growth bounds, the positive-support transform is analytic above a horizontal line and the negative-support transform below a horizontal line. If both functions have suitable subexponential growth, the basic domains are respectively the upper and lower half-planes; exponential decay can enlarge these to overlapping strips. The convolution theorem gives $(\bar{K}-1)\bar{f}=\bar{h}$.

The Wiener-Hopf method factors $\bar{K}-1$ into functions analytic and nonzero in the respective domains. After multiplicative or additive separation, the two sides continue to one entire function, whose growth determines a polynomial by Liouville's theorem. Real-axis zeros must be assigned consistently with the inversion contour, rather than divided out without specifying a side.

For $K=e^{-|x|}$, direct integration gives $\bar{K}=2/(1+k^2)$, soRearrange the Wiener-Hopf factorization asThe left side has the positive-support analytic continuation and the right side the negative-support one. Their $O(1)$ growth, from the supplied $O(k^{-1})$ bounds, makes the common continued function a constant $C$. ConsequentlyFor this kernel, $h$ on the negative axis is a multiple of $e^x$, so the latter transform is analytic for $\mathrm{Im}k<1$. The common strip can therefore be chosen as $0<\mathrm{Im}k<1$. In particular the inversion line must pass above both real poles $k=\pm 1$ of $\bar{f}$.

Set $C=iA$. ThenThis is the transform, on $\mathrm{Im}k>0$, ofEquivalently, Fourier inversion closed below the inversion line for $x>0$ picks up the two real poles with clockwise orientation.

As a direct check, let $G(x)={\int }_0^{\infty }{e}^{-|x-t|}f(t)dt$. Since $(d^2/d{x}^2)e^{-|x|}=e^{-|x|}-2{\delta }_0$, the equation $G=f$ gives $f^{\prime \prime }=-f$. At zero, differentiation of the integral gives $G^{\prime }(0)=G(0)$, so $f^{\prime }(0)=f(0)$; this sets the sine and cosine coefficients equal. Direct substitution, or the transforms above, then verifies the full integral equation, not merely its differentiated form.