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Now consider formula (2) below derived from the integral $f(0)=\int_0^{\infty}\delta(x)\ f(x)\, dx$ where $f(x)=e^{-\left| x\right|}$ and formula (1) above for $\delta(x)$ was used to evaluate the integral. Formula (2) below can also be evaluated as illustrated in formula (3) below.

Now consider formula (2) below derived from the integral $f(0)=\int_{-\infty}^{\infty}\delta(x)\ f(x)\, dx$ where $f(x)=e^{-\left| x\right|}$ and formula (1) above for $\delta(x)$ was used to evaluate the integral. Formula (2) below can also be evaluated as illustrated in formula (3) below.

The conditional convergence requirement $M(N)=0$ stated for formulas (1) to (7) above is because the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ defined in formula (7) above only evaluates to zero at $x=0$ when $M(N)=0$. The condition $M(N)=0$ is required when evaluating formula (7) above and formulas derived from the two Mellin convolutions defined in the preceding paragraph using formula (7) above, but I'm not sure it's really necessary when evaluating formula (1) above or formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ using formula (1) above (e.g. formulas (4), (5), and (6) above). Formula (1) above is based on the evaluation of formula (7) above at $|x|\ge 1$, so perhaps formula (1) above is not as sensitive to the evaluation of formula (7) above at $x=0$. Formula (1) above can be seen as taking formula (7) above, cutting out the strip $-1<x\le 1$, and then gluing the two remaining halves together at the origin. Nevertheless I usually evaluate formula (1) above and formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ using formula (1) above at $M(N)=0$ since it doesn't hurt anything to restrict the selection of $N$ to this condition and I suspect this restriction may perhaps lead to faster and/or more consistent convergence.

The conditional convergence requirement $M(N)=0$ stated for formulas (1) to (7) above is because the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ defined in formula (7) above only evaluates to zero at $x=0$ when $M(N)=0$. The condition $M(N)=0$ is required when evaluating formula (7) above and formulas derived from the two Mellin convolutions defined in the preceding paragraph using formula (7) above, but I'm not sure it's really necessary when evaluating formula (1) above or formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ using formula (1) above (e.g. formulas (4), (5), and (6) above). Formula (1) above is based on the evaluation of formula (7) above at $|x|\ge 1$, so perhaps formula (1) above is not as sensitive to the evaluation of formula (7) above at $x=0$. Formula (1) above can be seen as taking formula (7) above, cutting out the strip $-1\le x<1$, and then gluing the two remaining halves together at the origin. Nevertheless I usually evaluate formula (1) above and formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ using formula (1) above at $M(N)=0$ since it doesn't hurt anything to restrict the selection of $N$ to this condition and I suspect this restriction may perhaps lead to faster and/or more consistent convergence.