Calculate $\lim\limits_{R\rightarrow\infty}\int_0^\pi \cos(R\cos t)dt$ w.o. Bessel function Im calculating a complex path integral to calculate $\int_0^\infty \frac{\sin x}{x}dx$. I was able to evaluate everything except the arc $\int_0^\pi i~\exp(iR~e^{it})dt$ where $R$ is the radius.
I managed to use the $\lim\limits_{R\rightarrow \infty}$ to break the problem down to calculating $\lim\limits_{R\rightarrow \infty}\int_0^\pi i\cos(R \cos(t))dt$.
I know it evaluates to $\pi$ but we haven't introduced the Bessel function, which solves the integral. Is ther any other way to solve it? 
 A: Not a strict limit, but leading-order behavior in this limit.
Consider
$$I(R) = 2 \int_0^{\pi/2} dt \, e^{i R \cos{t}} $$
We may use the method of stationary phase: as $R \to \infty$, the value of the integral is dominated by the neighborhoods about which the derivative of the exponent vanishes, i.e., $\sin{t}=0$ or $t=0$.  The reason for this is, at the very high frequencies in this limit, the value of the integral tends to cancel by the Riemann-Lesbegue lemma.  Thus, we expand out to second order about a small neighborhood near this stationary point and get
$$I(R) \approx 2 \int_0^{\epsilon} dt \, e^{i R (1-t^2/2)} $$
We may show that, with small error, we may expand out the outer integration limit to infinity.  Thus,
$$I(R) \approx 2 e^{i R} \int_0^{\infty} dt \, e^{-i R t^2/2} =  e^{i R} \sqrt{\frac{2\pi}{i R}} =  \sqrt{2 \pi} e^{i (R-\pi/4)} R^{-1/2}$$
Noting that we only want the real part of the above expression, we have
$$I(R) \approx \sqrt{\frac{2 \pi}{R}} \cos{\left (R-\frac{\pi}{4} \right )} $$
A: Regarding your first integral, the usual way to do this is to note that
$$ \lvert e^{iR e^{it}} \rvert = \lvert e^{iR (\cos{t}+i\sin{t})} \rvert = e^{-R \sin{t}}, $$
because the absolute value of the exponential comes from the real part of the argument. You also have the inequality $\sin{t}\geqslant 2t/\pi$ for $0<t<\pi$, so
$$  \lvert e^{iR e^{it}} \rvert \leqslant e^{-2Rt/\pi}. $$
Now use the integral triangle inequality, $\lvert \int_a^b f \rvert < \int_a^b \lvert f \rvert $ to show that
$$ \left\lvert \int_0^{\pi/2} e^{iR e^{it}} \, dt \right\lvert \leqslant \int_0^{\pi} e^{-2Rt/\pi} \, dt \to 0 $$
as $R \to \infty$. The absolute value of the integrand is symmetric about $\pi/2$, so this is sufficient to show that the whole integral tends to zero.
