How Do You Evaluate Complex Contour Integrals with Sinusoidal Functions?

[riemannian]

greetings , we have the following integral :

$$I(x)=\lim_{T\rightarrow \infty}\frac{1}{2\pi i}\int_{\gamma-iT}^{\gamma+iT}\sin(n\pi s)\frac{x^{s}}{s}ds$$

n is an integer . and $\gamma >1$

if $x>1$ we can close the contour to the left . namely, consider the contour :

$$C_{a}=C_{1}\cup C_{2}\cup C_{3}\cup C_{4}$$

where :

$$C_{1}=\left [ \gamma-iT,\gamma+iT \right ]$$

$$C_{2}=\left [ \gamma+iT,-U+iT \right ]$$

$$C_{3}=\left [ -U+iT ,-U-iT \right ]$$

$$C_{4}=\left [ -U-iT ,\gamma-iT \right ]$$

and $U>>\gamma$
then by couchy's theorem :

$$I(x)=I_{1}+I_{2}+I_{3}+I_{4}=0$$

if $x<1$, we can close the contour to the right via the following contour :

$$C_{b}=C_{1}\cup C_{2}\cup C_{3}\cup C_{4}$$

$$C_{1}=\left [ \gamma-iT,\gamma+iT \right ]$$

$$C_{2}=\left [ \gamma+iT,U+iT \right ]$$

$$C_{3}=\left [ U+iT ,U-iT \right ]$$

$$C_{4}=\left [ U-iT ,\gamma-iT \right ]$$

then also by couchy's theorem :

$$I(x)=0$$

the plan is to give an estimate of the integrals along the segments of the rectangular contour, and calculate $I_{1}$ in both cases via the result obtained by cauchy's theorem . however, i don't have the first clue on how to do that, hence the quest !

 

[haruspex]

then by couchy's theorem :
$$I(x)=I_{1}+I_{2}+I_{3}+I_{4}=0$$

I don't think you mean that. Yes, by Cauchy, $$I_{1}+I_{2}+I_{3}+I_{4}=0$$, but the challenge is to show that three of these tend to 0 as T tends to infinity, and therefore the fourth (the one that tends to I(x)) does too.
I got stuck trying to put some bound on sin(s). When s = c+id has a large imaginary component, id, this grows like exp(d). I tried pairing up the negative imaginary points with the positive ones, e.g. doing C3 as the integral from -U to -U+iT of {f(s) - f(complex conjugate of s)}.dt, hoping to get some cancellation. No luck.

 

[mmzaj]

I don't think you mean that. Yes, by Cauchy, $$I_{1}+I_{2}+I_{3}+I_{4}=0$$, but the challenge is to show that three of these tend to 0 as T tends to infinity, and therefore the fourth (the one that tends to I(x)) does too.
I got stuck trying to put some bound on sin(s). When s = c+id has a large imaginary component, id, this grows like exp(d). I tried pairing up the negative imaginary points with the positive ones, e.g. doing C3 as the integral from -U to -U+iT of {f(s) - f(complex conjugate of s)}.dt, hoping to get some cancellation. No luck.

yes, sorry .. it was a typo . and i was trying to do exactly the same , no luck

remark : the integral of question bears resemblance to the perron's integral, and could be thought of as a laplace inverse of $\frac{\sin(n\pi s)}{s}$, with $\ln x$ as the variable in the 'time' domain.