Find the leading order of the integral $\int_0^1 \cos\left(x{t^4}\right)\tan t\,dt$, while $x$ goes to $+\infty$. I would like some help with this problem. I don't know where to start. Is it right to try integration by parts? Another method that I've learned is to find the Taylor series of each function and then integrate. But I am not sure if this is the right way to do it.
Thanks in advance. 
 A: The saddle point method is applicable. The steepest descent direction at $t = 0$ is $\phi = e^{i \pi/8}$. Taking $t = \phi u^{1/4}$ and expanding $t'(u) \tan t$ and $e^{i x t^4}$ around $u = 0$ gives
$$\int_0^{\phi \infty} e^{i x t^4} \tan t \, dt =
\frac \phi 4 \int_0^\infty \frac {e^{-x u} \tan \phi u^{1/4}} {u^{3/4}} du =
\frac \phi 4 \int_0^\infty e^{-x u}
 \left( \frac \phi {\sqrt u} + \frac {\phi^3} 3 \right) du + o(x^{-1}).$$
At $t = 1$, we take $t = 1 - i u$:
$$\int_{1 + i \infty}^1 e^{i x t^4} \tan t \, dt =
-i e^{i x} \tan 1 \int_{-\infty}^0 e^{4 x u} du + o(x^{-1}).$$
Taking the real part gives two terms in the asymptotic expansion of $\int_0^1 \cos(x t^4) \tan t \, dt$.
A: Substitute $u=xt^4$ to obtain
\begin{eqnarray}
\int_0^1\cos\left(xt^4\right)\tan t\mathrm dt
&=&
\frac14\int_0^x\cos u\tan\left(\left(\frac ux\right)^\frac14\right)\frac{\mathrm du}{x^{\frac14}u^{\frac34}}
\\
&=&
\frac14\int_0^x\cos u\left(\sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n - 1} 2^{2 n} \left({2^{2 n} - 1}\right) B_{2 n} \, \left(\left(\frac ux\right)^\frac14\right)^{2 n - 1} } {\left({2 n}\right)!}\right)\frac{\mathrm du}{x^{\frac14}u^{\frac34}}
\\
&=&
\frac14\int_0^x\left(\frac1{\sqrt x\sqrt u}\cos u+\frac13\frac1x\cos u+\cdots\right)\mathrm du
\\
&=&\frac14\left(\frac2{\sqrt x}\int_0^{\sqrt x}\cos^2u\,\mathrm du+\frac13\frac{\sin x}x+\cdots\right)
\\
&=&
\frac{C\left(\sqrt x\right)}{2\sqrt x}+\frac1{12}\frac{\sin x}x+\cdots\;,
\end{eqnarray}
where
$$
C(x)  =\int_0^x \cos t^2\,\mathrm dt
$$
is a Fresnel integral. (Note that there are different conventions for this integral; this is the Wikipedia definition; Wolfram|Alpha and MathWorld use a different convention.) With $\lim_{x\to\infty}C(x)=\sqrt{\frac\pi8}$, the leading term is
$$
\sqrt{\frac\pi{32x}}\;.
$$
