As $M \to +\infty$, how could I make a good asymptotic analysis of this integral?



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> $$\int\_0^1 \dfrac{\cos(M x)}{1 + x^2} e^{-M (x^2 - 1/9)}\ \text{d}x$$
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The exponential term shall dominate, yet I have no clue in who to deal with $\cos(Mx)$. I tried to apply geometric series for $\frac{1}{1+x^2}$ but an infinite sums might not be the best deal.


I am not sure if/when to use Taylor series. I thought about Laplace Method, but I think it could not work because of the $\cos(MX)$ function... Any hint?


Thank you!



Asymptotic expansion / analysis of this integral
For large $M$ the integrand contributes in the range $x\lesssim 1/\sqrt M$, so we can neglect the denominator $1+x^2$. The integral then has a closed form expression,
$$\int\_0^1 \cos (M x) e^{-M (x^2 - 1/9)}\ \text{d}x=\tfrac{1}{4}\sqrt{\pi }M^{-1/2}e^{M/9} e^{-M/4} \left[\text{erf}\left(\left(1+\tfrac{i}{2}\right) \sqrt{M}\right)-i \,\text{erfi}\left(\left(\tfrac{1}{2}+i\right) \sqrt{M}\right)\right].$$
The expression in square brackets tends to 2 in the limit $M\rightarrow\infty$, leaving us with the asymptotics
$$\int\_0^1 \dfrac{\cos M x}{1 + x^2} e^{-M (x^2 - 1/9)}\ \text{d}x\rightarrow \tfrac{1}{2}\sqrt{\pi }M^{-1/2}e^{M/9} e^{-M/4}.$$


