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I am trying to deal with the integral $$\int_0^{\pi/4}\cos nt \;e^{x\cos t} \; dt \quad x \rightarrow \infty$$ I want its asymptotic expansion, to do so I have split the integral into two parts, $$\int_0^{\epsilon}\cos nt \;e^{x\cos t} \; dt$$,$$\int_\epsilon^{\pi/4}\cos nt \;e^{x\cos t} \; dt$$

I would like to show that the second one is really small, $$\int_\epsilon^{\pi/4}\cos nt \;e^{x\cos t} \; dt \leq {\color{black}e^x \int_\epsilon^{\pi/4} \;e^{-\frac{1}{2!}xt^2+\frac{1}{4!}xt^4+...+...} \; dt} \leq$$ $$ \exp\bigg(\sum_i x\frac{1}{(4i)!} \bigg)\exp\bigg(\sum_i -x\frac{\epsilon^{4i+2}}{(4i+2)!} \bigg)\gamma$$

Note that I have bounded $t$ above by 1 and in the second by$\epsilon$. Since $\epsilon$ is small and $x >> 1$:

(1) $\epsilon^{4i+2} < \epsilon^{2}$, $i > 1$ and (2) $e^{-\epsilon x} < e^{\epsilon}$, $x > 1$

so using (1) $$\leq \exp\bigg(\sum_i x \bigg(\frac{1}{(4i)!} - \frac{\epsilon^2}{(4i+2)!} \bigg) \bigg) \gamma$$ Or I could say, using (1) and (2) $$\leq \exp\bigg(\sum_i \bigg(\frac{x}{(4i)!} - \frac{\epsilon^2}{(4i+2)!} \bigg) \bigg) \gamma$$

How can I convince myself that this integral is exponentially small as x gets very large? Or better still in the solutions to this this problem is assumes that $\epsilon^2 x >>1$, how can I deduce this condition from above?

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  • $\begingroup$ why should it be very small? you have $e^{x\cos t}$. Since $\cos t$>0, this exponential gets very large as $x\to \infty$. $\endgroup$ – Marcel May 23 '17 at 14:38
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A stationary exponent analysis goes as follows. The integral should be dominated by the stationary points of $x\cos(t)$. These are solutions to $\sin(t)=0$ which are of the form $t=k\pi$ for integer $k$.

The only such point covered in the integration range is $t=0$. So you approximate $\cos(t)\approx 1-t^2/2$ to get

$$\int_0^{\infty} e^{x(1-t^2/2)}=e^x\sqrt{\frac{\pi}{2x}}$$

(the prefactor $\cos(nt)$ you just evaluate at $t=0$)

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