In Bender and Orszag's Advanced Mathematical Methods for Scientists and Engineers, Exercise 6.56 (c) asks to find the leading behaviour of the integral $ \int_0^1 \cos(x t^4)\tan t\;\mathrm{d}t$ as $x\rightarrow\infty$ by using the method of stationary phase. Can anyone help?
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Write the integral as $\operatorname {Re} \int_0^1 e^{i x t^4} \tan t , dt$ and take $u = t^2$. – Maxim Feb 22 '21 at 12:06
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At $x\to\infty$ the main contribution to the integral comes from the region of $t<<1$, so you may simplify the integrand accordingly ($tan{t}\approx{t}$) – Svyatoslav Feb 22 '21 at 12:27
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@Maxim thanks for your message. why would you make the substitution $u=t^2$? – user157226 Feb 27 '21 at 14:20
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@Svyatoslav would your approach simply leave us to evaluate $\Re \int_0^\infty t e^{ixt^4}\mathrm{d}t$? – user157226 Feb 27 '21 at 14:21
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Yes, this is correct – Svyatoslav Feb 27 '21 at 14:24
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@Svyatoslav, however once one makes the substitution $u=t^2$, you get something like $\Re \frac12 \int_0^1 e^{ixu^2} du$. Is it sensible? How to proceed from there? – user157226 Feb 27 '21 at 17:44
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Make change $t=\sqrt{x}u$, then you can expand integration over $t$ from $(0,\sqrt{x})$ to $(0,\infty)$ (it is valide for getting the main asymptotics term). Integral $ \frac{1}{2\sqrt{x}} \int_0^\infty e^{it^2} dt $ can be evaluated in the closed form; then take the real part of it. – Svyatoslav Feb 27 '21 at 18:21
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https://math.stackexchange.com/questions/3999803/complex-gaussian-integral-by-change-of-variable/4000082#4000082 – Svyatoslav Feb 27 '21 at 18:24