Return to Answer

In physics we use the Sokhotski–Plemelj theorem to evaluate integrals from zero to infinity of cosine and sine in the form:

$$F(k)=\int_0^\infty (\cos kx +i\sin kx)\,dx=\pi\delta(k)+i\,{\cal P}\frac{1}{k}$$

where $\delta(k)$ is the Dirac delta function and ${\cal P}$ is a reminder that when the right-hand-side is integrated over $k$ one should take the Cauchy principal value of the integral: $$\int_{-\infty}^\infty F(k)g(k)\,dk=\pi g(0)+i\,{\cal P}\,\int_{-\infty}^\infty \frac{g(k)}{k}\,dk.$$

Incidentally, from this I would indeed associate $\int_0^\infty \sin x\,dx$ with $+1$, as in the OP.

In physics we use the Sokhotski–Plemelj theorem to evaluate integrals from zero to infinity of cosine and sine in the form:

$$F(k)=\int_0^\infty (\cos kx +i\sin kx)\,dx=\pi\delta(k)+i\,{\cal P}\frac{1}{k}$$

where $\delta(k)$ is the Dirac delta function and ${\cal P}$ is a reminder that when the right-hand-side is integrated over $k$ one should take the Cauchy principal value of the integral: $$\int_{-\infty}^\infty F(k)g(k)\,dk=\pi g(0)+i\,{\cal P}\,\int_{-\infty}^\infty \frac{g(k)}{k}\,dk.$$

Incidentally, from $F(1)=i$ I would indeed associate $\int_0^\infty \cos x\,dx$ and $\int_0^\infty \sin x\,dx$ with $0$ and $1$, respectively, as in the OP.

In physics we use the Sokhotski–Plemelj theorem to evaluate integrals from zero to infinity of cosine and sine in the form:

$$F(k)=\int_0^\infty (\cos kx +i\sin kx)\,dx=\pi\delta(k)-i\,{\cal P}\frac{1}{k}$$

where $\delta(k)$ is the Dirac delta function and ${\cal P}$ is a reminder that when the right-hand-side is integrated over $k$ one should take the Cauchy principal value of the integral: $$\int_{-\infty}^\infty F(k)g(k)\,dk=\pi g(0)-i\,{\cal P}\,\int_{-\infty}^\infty \frac{g(k)}{k}\,dk.$$

Incidentally, from this I would associate $\int_0^\infty \sin x\,dx$ with $-1$ rather than $+1$, as in the OP.

In physics we use the Sokhotski–Plemelj theorem to evaluate integrals from zero to infinity of cosine and sine in the form:

$$F(k)=\int_0^\infty (\cos kx +i\sin kx)\,dx=\pi\delta(k)+i\,{\cal P}\frac{1}{k}$$

where $\delta(k)$ is the Dirac delta function and ${\cal P}$ is a reminder that when the right-hand-side is integrated over $k$ one should take the Cauchy principal value of the integral: $$\int_{-\infty}^\infty F(k)g(k)\,dk=\pi g(0)+i\,{\cal P}\,\int_{-\infty}^\infty \frac{g(k)}{k}\,dk.$$

Incidentally, from this I would indeed associate $\int_0^\infty \sin x\,dx$ with $+1$, as in the OP.

In physics we use the Sokhotski–Plemelj theorem to evaluate integrals from zero to infinity of cosine and sine in the form:

$$F(k)=\int_0^\infty (\cos kx +i\sin kx)\,dx=\pi\delta(k)-i\,{\cal P}\frac{1}{k}$$

where $\delta(k)$ is the Dirac delta function and ${\cal P}$ is a reminder that when the right-hand-side is integrated over $k$ one should take the Cauchy principal value of the integral: $$\int_{-\infty}^\infty F(k)g(k)\,dk=\pi g(0)-i\,{\cal P}\,\int_{-\infty}^\infty \frac{g(k)}{k}\,dk.$$

In physics we use the Sokhotski–Plemelj theorem to evaluate integrals from zero to infinity of cosine and sine in the form:

$$F(k)=\int_0^\infty (\cos kx +i\sin kx)\,dx=\pi\delta(k)-i\,{\cal P}\frac{1}{k}$$

where $\delta(k)$ is the Dirac delta function and ${\cal P}$ is a reminder that when the right-hand-side is integrated over $k$ one should take the Cauchy principal value of the integral: $$\int_{-\infty}^\infty F(k)g(k)\,dk=\pi g(0)-i\,{\cal P}\,\int_{-\infty}^\infty \frac{g(k)}{k}\,dk.$$

Incidentally, from this I would associate $\int_0^\infty \sin x\,dx$ with $-1$ rather than $+1$, as in the OP.