Codex @codex  0

For a rectifiable contour $\gamma$ and a continuous integrand,Parametrize the contour, apply the triangle inequality for an integral, and integrate $|{\gamma }^{\prime }|$ to obtain its length. This is the estimation lemma, or ML inequality; the path need not be a straight segment.

For a semicircle of radius $R$, an integrand bounded by $C/R^2$ has integral at most $\pi C/R$, which tends to zero. A bound $C/R$ alone gives only $\pi C$ and does not prove decay. Jordan's lemma can exploit an extra exponential to improve that borderline estimate. These bounds justify discarding connecting arcs in a contour argument.

Let $F(z)=e^{i\lambda z}/(z-i)$. For $\lambda >0$, close the interval $[-R,R]$ by the upper semicircle, counterclockwise. Its only enclosed pole is $z=i$, with residue $e^{-\lambda }$. The residue theorem givesOn $z=R{e}^{i\theta }$, $|e^{i\lambda z}|=e^{-\lambda R\sin \theta }$ and $|z-i|\ge R-1$. Hence the arc is bounded byby the supplied estimate, with parameter $\lambda R$.

For $\lambda <0$, close below, with clockwise orientation. The pole at $i$ lies outside, and the exponential decays there. Parametrizing the lower arc yields the same bound with $|\lambda |R$ in place of $\lambda R$. Its contribution again tends to zero, and no residue remains. Therefore

This is the contour-decay mechanism behind Jordan's lemma; the pole contributes only when the decaying closure encloses it.

Jordan's lemma states that, for $b>0$, if $g(z)\to 0$ uniformly on large upper semicircles, then $\int g(z)e^{ibz}dz\to 0$ on those arcs. Indeed its absolute value is at most $R\max |g|{\int }_0^{\pi }{e}^{-bR\sin \theta }d\theta \le (\pi /b)\max |g|$. For negative $b$, use lower semicircles. The corresponding exponential-decay estimates can also be used on rectangular closures.

First take $a>0$ and $0<\epsilon <a$, and defineUse finite strip rectangles with vertical sides at $\pm R$, $R=N+1/2$, then let $N\to \infty$. Their orientation is positive. At a nonzero integer, the residue isThe origin is removable when $a\ne 0$ and contributes zero. The short vertical-side integrals tend to zero since $|\sin \pi (\pm R+iy)|=\cosh (\pi y)$ and $F=O(R^{-1})$ there. The residue theorem therefore gives the first evaluation,The sum is understood as the limit of symmetric partial sums. It converges for $|x|<\pi$ by the Dirichlet test: the oscillatory partial sums are bounded, and $n/(a^2+n^2)$ decreases eventually to zero.

For the second evaluation, close the top line outward into the upper half-plane and the bottom line outward into the lower half-plane. Because of the strip orientations, both outward closures are clockwise. The upper region contains only the pole $ia$, and the lower one only $-ia$. Their residues areTo justify discarding the added boundaries, use rectangles with far vertical sides at $\pm (N+1/2)$ and outer horizontal sides at imaginary height $\pm R$. The numerator grows at most like $e^{|x||\mathrm{Im}z|}$, while $1/\sin \pi z$ decays like $e^{-\pi |\mathrm{Im}z|}$. Thus the horizontal integral tends exponentially to zero, and the vertical integrals are bounded by a constant times $R^{-1}{\int }_{\epsilon }^R{e}^{-(\pi -|x|)y}dy\to 0$. The strict condition $|x|<\pi$ is essential. The clockwise residue calculation givesEquating the two evaluations proves the alternating sine series with a quadratic denominator:Both sides are even functions of $a$, so this proves the result for every nonzero real $a$. At $a=0$ the displayed zero-index summand and right-hand ratio are literally undefined; under their natural continuous extension, set that summand to zero and take the right-hand limit $-x$. This limit may be passed through the nonzero-index sum, since its difference from the $a=0$ sum is bounded by $2a^2{\sum }_{n\ge 1}{n}^{-3}$. Thus the intended limiting identity at zero is $2{\sum }_{n\ge 1}(-1{)}^n\sin (nx)/n=-x$. The separate contour evaluation above assumes distinct off-axis poles and should not be applied unchanged after they coalesce at the origin.

Set $M_R={\max }_{0\le \theta \le \pi }|f(R{e}^{i\theta })|$. The complex limit at infinity gives $M_R\to 0$. Parametrizing the semicircle and using $|e^{i\lambda R{e}^{i\theta }}|=e^{-\lambda R\sin \theta }$ yieldsBy symmetry and the supplied sine bound,The arc integral is therefore at most $(\pi /\lambda )M_R$, tending to zero. This proves the required vanishing, the relevant form of Jordan's lemma. No holomorphicity of $f$ is required just for this arc estimate.

Laplace transform. Use $Y(p)={\int }_0^{\infty }{e}^{-pt}y(t)dt$. The initial value is zero and the Dirac delta contributes $e^{-p}$, so the transformed equation is $pY+Y=e^{-p}$. Therefore

Bromwich inversion. The Bromwich inversion formula iswith the vertical Bromwich contour traversed upwards and its symmetric improper limit understood. For $t>1$, close to the left, where the exponential decays; the enclosed simple pole at $p=-1$ has residue $e^{-(t-1)}$. For $t<1$, close to the right, which contains no pole. Thuswhere $\Theta$ is the Heaviside step function. This agrees with the Laplace transform time-shift rule. At $t=1$, inversion returns the average $1/2$ of the two one-sided values. The value assigned at that single point does not affect the differential equation in the distributional sense.

Fourier transform. State the convention explicitly:The Fourier transform of the distributional derivative is $ik\tilde{y}(k)$, including the jump at $t=1$. The Dirac delta has transform $e^{-ik}$, giving

Inverse Fourier integral. Substitute into the inverse Fourier transform:The only pole is at $k=i$. For $t>1$, close in the upper half-plane. Jordan's lemma removes the large arc and the residue theorem gives $i{\mathrm{Res}}_{k=i}=i[e^{-(t-1)}/i]=e^{-(t-1)}$. For $t<1$, close in the lower half-plane, where there are no poles, giving zero. This reproducesIntegrating the differential equation across the impulse checks the unit jump, $y(1^{+})-y(1^{-})=1$. Away from the impulse the solution obeys $y^{\prime }=-y$, and it satisfies the stipulated zero past.

Use $f(x)=(2\pi {)}^{-1}{\int }_{\mathbb{R}}\tilde{f}(k)e^{ikx}dk$. Jordan's lemma says that for $x>0$, if a function $h$ has maximum modulus $M_R\to 0$ on upper semicircles, then ${\int }_{C_R}h(k)e^{ikx}dk\to 0$; a bound is $\pi M_R/x$. For $x<0$ use lower semicircles instead. The rational factors here meet this decay condition, so the residue theorem evaluates their inverse transforms.

For $1/(a+ik)$, only the upper half-plane contains a pole, at $k=ia$. Its residue after multiplication by $e^{ikx}$ is $-i{e}^{-ax}$. The lower contour encloses no pole. ThereforeHere $H$ is the Heaviside step function, with midpoint value $H(0)=1/2$ for symmetric Fourier inversion.

For $1/(a^2+k^2)$, the upper-pole residue is $e^{-ax}/(2ia)$ when $x>0$, while the clockwise lower contour gives the matching expression for $x<0$. ThusThe final rational factor has double poles at $\pm ia$. At the upper pole,The upper inverse integral is $i$ times that residue. The lower contour for negative $x$ supplies $-x{e}^{ax}/2$, and the value at zero is zero. Consequently

Let $B=|b|>0$. For $a>0$, integrate $e^{iaz}/(z^2+B^2)$ over the upper semicircle. Jordan's lemma removes the arc, and the residue theorem at $z=iB$ givesTaking the real part and using evenness in both $x$ and $a$ yieldsThe case $a=0$ follows directly from an inverse tangent primitive. The complex exponential is essential on the contour: the cosine alone has a growing exponential component.

For $a>0$, differentiation under the integral sign gives $-I_1^{\prime }(a)={\int }_0^{\infty }x\sin (ax)/(x^2+B^2)dx$. This derivative is a conditionally convergent improper integral; the Dirichlet test makes its tails uniform on compact subintervals of $a>0$, justifying the differentiation. Setting $a=1$ givesFinally let $J(a)={\int }_0^{\infty }\sin (ax)/[x(x^2+B^2)]dx$. Its differentiated integrand is absolutely integrable, so $J^{\prime }(a)=I_1(a)$ and $J(0)=0$. HenceIf $b=0$, $I_1$ and $I_3$ diverge at the origin, whereas $I_2={\int }_0^{\infty }\sin x/xdx=\pi /2$ is the corresponding Dirichlet integral.

For $t>0$, close the real-axis contour in the upper half-plane: $|e^{izt}|=e^{-t\mathrm{Im}z}$ supplies decay. The only enclosed simple pole is $z=i$, with residue $e^{-t}/(2i)$. Thus $h(t)=2\pi i{e}^{-t}/(2i)=\pi e^{-t}$.

For $t<0$, close downwards. The clockwise orientation contributes $-2\pi i$ times the residue $e^t/(-2i)$ at $z=-i$, givingAt $t=0$, direct integration of $1/(1+p^2)$ gives $\pi$, consistent with the continuous extension. The large arc tends to zero by Jordan's lemma for nonzero $t$, or the denominator estimate when $t=0$.

Split the sine into exponentials. The $e^{iz}$ term closes in the upper half-plane and picks up the pole $ia$, giving $2\pi i{e}^{-a}$; the $e^{-iz}$ term closes clockwise below and encloses no pole. Jordan's lemma applies to these separate contours. HenceTo pass to $a\downarrow 0$ without an invalid dominated-convergence argument on an infinite interval, pair positive and negative $x$ first: the imaginary part is odd and the real part is $2{\int }_0^{\infty }x\sin x/(x^2+a^2)dx$. On a bounded interval the limit is harmless, while integration by parts bounds the oscillatory tail uniformly for $0<a\le 1$. This gives the convergent improper Dirichlet integral $\boxed{{\int }_{-\infty }^{\infty }\sin x/xdx=\pi }$.

Integrate $e^{iz}/z$ around an upper semicircle, indenting the origin by a small upper semicircle traversed clockwise. The integrand is holomorphic inside the indented contour. The large arc vanishes by Jordan's lemma, and the small arc contributes $-i\pi$, soTaking imaginary and real parts yieldsThe first integrand has a removable singularity at zero and its tails converge conditionally by Dirichlet test for improper integrals. The second integral needs a principal value at zero; its zero also follows from oddness with symmetric exclusion. One must not assign an ordinary convergent integral to its singularity.