Complex Analysis: Very Difficult Problems with Solutions

Solution:
The integrand is analytic except at the zeros of [tex]z^{3}\cos z[/tex], i.e. at [tex]z=0[/tex] and [tex]z=\frac{\pi}{2}+k\pi[/tex], [tex]k\in\mathbb{Z}[/tex]. Inside the circle lie only [tex]0[/tex] and [tex]\pm\frac{\pi}{2}\approx\pm 1.57[/tex], because [tex]\frac{3\pi}{2}\approx 4.71\gt 4[/tex]; none of these points lies on the circle.
At [tex]z=0[/tex] the numerator [tex]e^{z}-1[/tex] has a simple zero and [tex]\cos 0=1[/tex], so [tex]z=0[/tex] is a pole of order [tex]2[/tex]. Multiply the expansions
[tex]\frac{e^{z}-1}{z^{3}}=\frac{1}{z^{2}}+\frac{1}{2z}+\frac{1}{6}+\ldots,\qquad \frac{1}{\cos z}=1+\frac{z^{2}}{2}+\ldots[/tex]
Since [tex]\frac{1}{\cos z}[/tex] is even, it has no [tex]z^{1}[/tex] term, so the coefficient of [tex]\frac{1}{z}[/tex] in the product comes only from [tex]\frac{1}{2z}[/tex] multiplied by [tex]1[/tex]:
[tex]\operatorname{Res}_{z=0}=\frac{1}{2}[/tex]
At [tex]z=\pm\frac{\pi}{2}[/tex] the cosine has simple zeros ([tex](\cos z)'=-\sin z\ne 0[/tex]) and the numerator is non-zero, so these are simple poles and
[tex]\operatorname{Res}_{z=\pm\frac{\pi}{2}}=\left.\frac{e^{z}-1}{z^{3}(-\sin z)}\right|_{z=\pm\frac{\pi}{2}}[/tex]
[tex]\operatorname{Res}_{z=\frac{\pi}{2}}=\frac{e^{\frac{\pi}{2}}-1}{-\frac{\pi^{3}}{8}}=-\frac{8\left(e^{\frac{\pi}{2}}-1\right)}{\pi^{3}}[/tex]
[tex]\operatorname{Res}_{z=-\frac{\pi}{2}}=\frac{e^{-\frac{\pi}{2}}-1}{-\frac{\pi^{3}}{8}}=-\frac{8\left(e^{-\frac{\pi}{2}}-1\right)}{\pi^{3}}[/tex]
By the residue theorem
[tex]\oint_{|z|=4}\frac{e^{z}-1}{z^{3}\cos z}\,dz=2\pi i\left[\frac{1}{2}-\frac{8}{\pi^{3}}\left(e^{\frac{\pi}{2}}+e^{-\frac{\pi}{2}}-2\right)\right]=\pi i-\frac{16i}{\pi^{2}}\left(e^{\frac{\pi}{2}}+e^{-\frac{\pi}{2}}-2\right)\approx -1.752\,i[/tex]

Solution:
The integrand [tex]\frac{\sin x}{x(x^{2}+1)}[/tex] is continuous on [tex]\mathbb{R}[/tex] (its limit at [tex]0[/tex] is [tex]1[/tex]) and is [tex]O\left(\frac{1}{|x|^{3}}\right)[/tex] as [tex]|x|\to\infty[/tex], so the integral converges. Consider
[tex]f(z)=\frac{e^{iz}}{z(z^{2}+1)}[/tex]
with simple poles [tex]0[/tex] and [tex]\pm i[/tex]. Integrate it over the closed contour made of the segment [tex][-R, -\varepsilon][/tex], the small upper semicircle [tex]\gamma_{\varepsilon}[/tex] of radius [tex]\varepsilon[/tex] centered at [tex]0[/tex] and traversed clockwise, the segment [tex][\varepsilon, R][/tex] and the large upper semicircle [tex]\Gamma_R[/tex] ([tex]0\lt\varepsilon\lt 1\lt R[/tex]). Only the pole [tex]z=i[/tex] lies inside, and
[tex]\operatorname{Res}_{z=i}f=\frac{e^{-1}}{i(2i)}=-\frac{1}{2e},\qquad \oint f\,dz=2\pi i\left(-\frac{1}{2e}\right)=-\frac{\pi i}{e}[/tex]
On [tex]\Gamma_R[/tex] we have [tex]|e^{iz}|=e^{-\operatorname{Im}z}\le 1[/tex] and [tex]|z^{3}+z|\ge R^{3}-R[/tex], so
[tex]\left|\int_{\Gamma_R}f\,dz\right|\le\frac{\pi R}{R^{3}-R}\to 0\quad (R\to\infty)[/tex]
For a simple pole, the integral over a small clockwise semicircle around it tends to [tex]-\pi i[/tex] times the residue; here [tex]\operatorname{Res}_{z=0}f=1[/tex], so
[tex]\int_{\gamma_{\varepsilon}}f\,dz\to-\pi i\quad (\varepsilon\to 0)[/tex]
Letting [tex]\varepsilon\to 0[/tex] and [tex]R\to\infty[/tex], the integrals over the two segments give the principal value [tex]J[/tex]:
[tex]J=\lim_{\varepsilon\to 0}\left(\int_{-\infty}^{-\varepsilon}+\int_{\varepsilon}^{\infty}\right)\frac{e^{ix}}{x^{3}+x}\,dx,\qquad J-\pi i=-\frac{\pi i}{e}\ \Rightarrow\ J=\pi i\left(1-\frac{1}{e}\right)[/tex]
Since [tex]e^{ix}=\cos x+i\sin x[/tex] and [tex]\frac{\cos x}{x^{3}+x}[/tex] is odd, the real part of [tex]J[/tex] is [tex]0[/tex], while its imaginary part is the required integral (whose integrand is continuous at [tex]0[/tex]). Hence
[tex]\int_{-\infty}^{+\infty}\frac{\sin x}{x^{3}+x}\,dx=\operatorname{Im}J=\pi\left(1-\frac{1}{e}\right)\approx 1.986[/tex]

Solution:
Put [tex]\rho=\limsup_{k\to\infty}\sqrt[k]{|a_{-k}|}[/tex]. With [tex]\zeta=\frac{1}{z-a}[/tex] the principal part
[tex]P(z)=\sum_{k=1}^{\infty}a_{-k}(z-a)^{-k}=\sum_{k=1}^{\infty}a_{-k}\zeta^{k}[/tex]
becomes a power series in [tex]\zeta[/tex] with radius of convergence [tex]\frac{1}{\rho}[/tex] (Cauchy-Hadamard formula, with [tex]\frac{1}{0}=\infty[/tex]). Hence [tex]P[/tex] converges absolutely and uniformly on compact subsets of [tex]|z-a|\gt\rho[/tex] and diverges for [tex]0\lt|z-a|\lt\rho[/tex]. Since all [tex]\sqrt[k]{|a_{-k}|}\ge 0[/tex], the condition [tex]\lim_{k\to\infty}\sqrt[k]{|a_{-k}|}=0[/tex] is equivalent to [tex]\rho=0[/tex]. The regular part [tex]Q(z)=\sum_{k=0}^{\infty}a_{k}(z-a)^{k}[/tex] converges at every point of [tex]G[/tex], so its radius of convergence is at least [tex]R[/tex] and [tex]Q[/tex] is analytic in [tex]|z-a|\lt R[/tex].
Sufficiency. If [tex]\rho=0[/tex], then [tex]P[/tex] converges for every [tex]z\ne a[/tex], uniformly on compact subsets of [tex]|z-a|\gt 0[/tex], so by the Weierstrass theorem it is analytic there. Then [tex]g=P+Q[/tex] is analytic in [tex]0\lt|z-a|\lt R[/tex] and equals [tex]f[/tex] in [tex]G[/tex].
Necessity. Let such a function [tex]g[/tex] exist. Being analytic in the annulus [tex]0\lt|z-a|\lt R[/tex], it has a Laurent expansion converging there:
[tex]g(z)=\sum_{k=-\infty}^{\infty}b_{k}(z-a)^{k},\qquad b_{k}=\frac{1}{2\pi i}\oint_{|z-a|=s}\frac{g(z)}{(z-a)^{k+1}}\,dz,\quad 0\lt s\lt R[/tex]
Choosing [tex]r\lt s\lt R[/tex], where [tex]g=f[/tex], we get [tex]b_{k}=a_{k}[/tex] for all [tex]k[/tex] (uniqueness of the Laurent expansion of [tex]f[/tex] in [tex]G[/tex]). Hence [tex]\sum_{k=1}^{\infty}a_{-k}(z-a)^{-k}[/tex] converges for every [tex]0\lt|z-a|\lt R[/tex], i.e. the power series [tex]\sum_{k=1}^{\infty}a_{-k}\zeta^{k}[/tex] converges for all [tex]|\zeta|\gt\frac{1}{R}[/tex], so for arbitrarily large [tex]|\zeta|[/tex]. Its radius of convergence is therefore infinite, i.e. [tex]\rho=0[/tex], which is the required condition.