Compute the integrals \[ \int_{|z-2|=1} \frac{e^z}{z(z-1)^2}\,dz, \qquad \int_0^\infty \frac{\cos 2x}{x^2 + 2}\,dx. \]
Solution
The first integral is not defined as an ordinary contour integral, nor does its usual one-sided improper interpretation converge. The second integral is \[ \boxed{\frac{\pi}{2\sqrt2}e^{-2\sqrt2}}. \]
1. The first contour passes through a double pole,and its one-sided integral diverges there.
One has \(|1-2|=1\), so the pole at \(z=1\) lies on the path, not strictly inside it. Set \(g(z)=e^z/z\), which is holomorphic near \(1\). Direct calculation gives \(g(1)=e\) and \(g'(1)=0\). Taylor expansion therefore gives \[ \frac{e^z}{z(z-1)^2}=\frac{e}{(z-1)^2}+h(z) \] for a function \(h\) holomorphic near \(1\) [@SS03].
Parametrize the circle by \(z(t)=2+e^{it}\) for \(0\leq t\leq2\pi\). Then \(z(\pi)=1\) and \(|z'(t)|=1\). Choose \(\delta>0\) so the subarc with \(|t-\pi|\leq\delta\) is in the neighborhood where \(h\) is holomorphic and bounded. On its left portion, for \(0<\varepsilon<\delta\), the principal part integrates exactly to \[ \int_{\pi-\delta}^{\pi-\varepsilon} \frac{e\,z'(t)}{(z(t)-1)^2}\,dt =-\frac{e}{z(\pi-\varepsilon)-1} +\frac{e}{z(\pi-\delta)-1}. \] The first term has modulus tending to infinity, since its denominator tends to zero. The second term is fixed. The contribution of \(h(z(t))z'(t)\) has a finite limit because it is bounded and continuous on the closed subarc. Thus the one-sided integral of the full integrand cannot converge. In particular the displayed contour integral has no value in the ordinary sense. A zero residue at a double pole does not remove this obstruction.
2. The second integral has the displayed value.
Put \(F(z)=e^{2iz}/(z^2+2)\). For \(R>\sqrt2\), integrate over the interval \([-R,R]\) followed by the counterclockwise upper semicircle \(\Gamma_R\). The only pole inside is \(i\sqrt2\), and it is simple, with residue \(e^{-2\sqrt2}/(2i\sqrt2)\). The residue theorem gives \[ \int_{-R}^R\frac{e^{2ix}}{x^2+2}\,dx +\int_{\Gamma_R}F(z)\,dz =\frac{\pi}{\sqrt2}e^{-2\sqrt2} \] [@SS03]. On \(\Gamma_R\), one has \(|e^{2iz}|=e^{-2\operatorname{Im}z}\leq1\) and \(|z^2+2|\geq R^2-2\). Consequently \[ \left|\int_{\Gamma_R}F(z)\,dz\right| \leq\frac{\pi R}{R^2-2}\longrightarrow0. \] The real-axis integral converges absolutely, since its integrand has modulus \(1/(x^2+2)\). Letting \(R\to\infty\) and taking real parts yields \[ \int_{-\infty}^{\infty}\frac{\cos2x}{x^2+2}\,dx =\frac{\pi}{\sqrt2}e^{-2\sqrt2}. \] The integrand is even, so its half-line integral is half that value, as asserted.