At $x\to \infty $ approximate the integral
$$I(x)=\int\limits_{0}^{\infty }\cos \left ( x\left ( t^2-t^4 \right ) \right )dt$$
My attempt:
$$f(t) = t^2-t^4\Rightarrow f'(t) = 2t – 4t^3 = 2t(1 – 2t^2) = 0$$
I've tried to break it down in Taylor's row
$$
\cos(xf(t))\sim \cos\left(\frac{x}{\sqrt{2}}\right)-\frac{x}{\sqrt{2}}f'(t_1)\sin\left(\frac{x}{\sqrt{2}}\right)\cdot (t-t_1)+O((t-t_1)^2).
$$
$$
\begin{aligned}
I(x)&\sim\int\limits_{0}^{\infty }\cos \left (\frac{x}{\sqrt{2}}\right)dt-\frac{x}{\sqrt{2}}f'(t_1)\int\limits_{0}^{\infty }(t-t_1)\sin\left(\frac{x}{\sqrt{2}}\right)(t^2-t^4)dt\\
&=\sqrt{\frac{\pi }{2}}\cos \left (\frac{x}{\sqrt{2}}\right)+\frac{x}{\sqrt{2}}f'(t_1)\left(\frac{3}{x^3}-\frac{15}{x^5}\right)+O\left(\frac{1}{x^6}\right)\\
&=\sqrt{\frac{\pi }{2}}\cos \left (\frac{x}{\sqrt{2}}\right)+\frac{3\sqrt{2}}{2x}+\frac{15\sqrt{2}}{4x^3}+O\left(\frac{1}{x^5}\right).
\end{aligned}
$$
Have I made the right decision? If so, what should I do next?
Best Answer
For positive values of $x$ $$I(x)=\int\limits_{0}^{\infty }\cos \left ( x\left ( t^2-t^4 \right ) \right )\,dt$$is such that $$\frac{4 \sqrt{2}}{\pi }\,I(x)=\sin \left(\frac{x+3 \pi}{8} \right) J_{-\frac{1}{4}}\left(\frac{x}{8}\right)+\cos \left(\frac{x+\pi }{8}\right) J_{\frac{1}{4}}\left(\frac{x}{8}\right)$$ (have a look here)
Expanding the trignonometric functions and later using double angle formulae, for large value of $x$ $$\large\color{blue}{I(x)=\frac{1}{2} \sqrt{\frac{\pi}{x}}\left( P(x)\,\cos \left(\frac{x}{4}\right)+Q(x)\,\sin \left(\frac{x}{4}\right)+\frac 1 {\sqrt 2}P(x)\right)}$$
where $$P(x)=1-\frac{3}{4 x}-\frac{105}{32 x^2}+\frac{3465}{128 x^3}+O\left (\frac {1} {x^4} \right)$$ $$Q(x)=1+\frac{3}{4 x}-\frac{105}{32 x^2}-\frac{3465}{128 x^3}+O\left (\frac {1} {x^4} \right)$$
This is a very good approximation even for quite low value of $x$. For example, using $x=100$, the exact value is $0.13752183$ while the above approximation gives $\color{red}{0.137521}47$.