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I'm trying to use the residue calculus to evaluate

$$\oint_\Gamma \cos(\log|z|)\cosh(\text{Arg}(z))\text{Arg}(z)e^{is(z-1)}dz,$$ where $s>0$, and where $\text{Arg}$ is the principal argument, $\Gamma=\gamma_1+\gamma_2+\gamma_3$ is the contour with: $$\gamma_1: z=1+i0\to z=1+iR,$$ $$\gamma_2: z=1+Re^{i\theta},\text{ where }\theta=\pi/2\to 0,$$ and $$\gamma_3:z=x,\text{ where }x=R\to 1,$$ and eventually taking $R\to\infty$.

My try: I can see there are branch cuts here. For the log term I can choose a branch cut along the negative real axis, uncluding the origin, and for the $\text{Arg}$ function I can choose the line from $i$ to $-i$. None of these branch cuts seems to affect my chosen contour, so no worries there (so long as I'm right about my branch cuts that is)

Substituting $z=1+Re^{i\theta}$ for $\gamma_2$ gives $$-\int_0^\infty\cos\left(\frac{1}{2}\log(1+R^2e^{i2\theta})\right)\cosh\left(\tan^{-1}\left(\frac{R\sin\theta}{1+R\cos\theta}\right)\right)\tan^{-1}\left(\frac{R\sin\theta}{1+R\cos\theta}\right)e^{isRe^{i\theta}}iRe^{i\theta}d\theta,$$ and taking the limit as $R\to\infty$ gives zero (according to Mathematica)

The integral along $\gamma_3$ is zero since the argument of a real number is zero. Hence, we get

$$I= \int_1^{1+i\infty}\cos(\log|z|)\cosh(\text{Arg}z)\text{Arg}(z)e^{is(z-1)}dz + 0 + 0 = \sum_{\text{residues}},$$ and since there are no poles then the integral is equal to zero.

But! Looking closer at this integral, using the substitution $z=1+iy$, we get

$$I=i\int_0^\infty\cos\left(\log\sqrt{1+y^2}\right)\cosh(\tan^{-1} y)\tan^{-1}(y)e^{-sy}dy.$$ If we substitute some real values for $s$, numerical integration shows this integral is definitely not $0$. What am I doing wrong?

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    $\begingroup$ Does the integrand of the original integral even look analytic? $\endgroup$ – Ron Gordon Feb 2 '14 at 17:45
  • $\begingroup$ @Ron come to think of it, I'm guessing not since, for example, the function $f(x+iy)\equiv\text{Arg}(x+iy)=\tan^{-1}(y/x)+i0$ is, by the C-R equations, only analytic on the line $x=0$ ($y\neq 0$). Is this along the lines you were thinking... $\endgroup$ – Antinous Feb 2 '14 at 19:28
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    $\begingroup$ Yes, exactly. $\,$ $\endgroup$ – Ron Gordon Feb 2 '14 at 20:37

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