How do I show that the function $$ y(x) = \int_0^\infty \cos(t^3/3 + xt)~dt\qquad -\infty \lt x \lt \infty,$$ satisfies the differential equation $y''=xy$?
I can't simply differentiate under integral since the integrand is oscillatory and does not decay as $t$ becomes large.
First we should note that the integral form solution of Airy's differential equation $y''-xy=0$ can be found by assuming a suitable integral kernel which is so-called the “integral kernel method”. It is a mutation of using integral transform. But it is better than using integral transform, since inverse transform should not be needed.
Let $y=\int_Ce^{xs}K(s)~ds$,
Then $(\int_Ce^{xs}K(s)~ds)''-x\int_Ce^{xs}K(s)~ds=0$
$\int_Cs^2e^{xs}K(s)~ds-\int_Ce^{xs}K(s)~d(xs)=0$
$\int_Cs^2e^{xs}K(s)~ds-\int_CK(s)~d(e^{xs})=0$
$\int_Cs^2e^{xs}K(s)~ds-[e^{xs}K(s)]_C+\int_Ce^{xs}~d(K(s))=0$
$\int_Cs^2e^{xs}K(s)~ds-[e^{xs}K(s)]_C+\int_Ce^{xs}K'(s)~ds=0$
$-[e^{xs}K(s)]_C+\int_C(K'(s)+s^2K(s))e^{xs}~ds=0$
$\therefore K'(s)+s^2K(s)=0$
$K'(s)=-s^2K(s)$
$\frac{K'(s)}{K(s)}=-s^2$
$\int\frac{K'(s)}{K(s)}~ds=\int-s^2~ds$
$\ln K(s)=-\frac{s^3}{3}+c_1$
$K(s)=ce^{-\frac{s^3}{3}}$
$\therefore y=\int_Cce^{-\frac{s^3}{3}+xs}~ds$
But since the above procedure in fact suitable for any complex number $s$,
$\therefore y_n=\int_{a_n}^{b_n}c_ne^{-\frac{((p_n+q_ni)t)^3}{3}+x(p_n+q_ni)t}~d((p_n+q_ni)t)$
$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{-\frac{(p_n^3+3p_n^2q_ni-3p_nq_n^2-q_n^3i)t^3}{3}+(p_n+q_ni)xt}~dt$
$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}~dt$
For some $x$-independent real number choices of $a_n$ , $b_n$ , $p_n$ and $q_n$ such that:
$\displaystyle\lim_{t\to a_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}=\lim_{t\to b_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}$
$\int_{a_n}^{b_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}~dt$ converges
For $n=1$, the best choice is $a_1=-\infty$ , $b_1=\infty$ , $p_1=0$ , $q_1=1$
$\therefore y_1=ic_1\int_{-\infty}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt$
$=ic_1\left(\int_{-\infty}^{0}e^{\left(\frac{t^3}{3}+xt\right)i}~dt+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$
$=ic_1\left(\int_{\infty}^{0}e^{\left(\frac{(-t)^3}{3}+x(-t)\right)i}~d(-t)+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$
$=ic_1\left(\int_{0}^{\infty}e^{\left(-\frac{t^3}{3}-xt\right)i}~dt+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$
$=ic_1\left(\int_{0}^{\infty}e^{-\left(\frac{t^3}{3}+xt\right)i}~dt+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$
$=C_1\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
“integral kernel method” not only suitable for Airy's differential equation, it is in fact at least suitable for many high order linear differential equation with linear coefficients such as $y''+y'+xy=0$ (http://tw.group.knowledge.yahoo.com/ignored-knowledgeunion/article/view?aid=4) , $y''+2y'-2xy=0$ (http://tw.knowledge.yahoo.com/question/article?qid=1711011401283) , $ y''+xy'+xy=0$ (http://tw.knowledge.yahoo.com/question/question?qid=1009100207053) , even for $ y'''-xy''-4y'+3y=0$ (http://tw.knowledge.yahoo.com/question/question?qid=1011072501482) and $y^{(n)}=axy+b$ (http://eqworld.ipmnet.ru/en/solutions/ode/ode0411.pdf) , etc.
For proving $\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$ satisfies the differential equation $y''-xy=0$ , we should highly aware the convergences of the improper integrals of the form $\int_{c}^{\infty}x^n\sin(P(x))~dx$ and $\int_{c}^{\infty}x^n\cos(P(x))~dx$, where $c$ and $n$ are real numbers, $P(x)$ is a $m$-degree polynomial ($m\geq 2$) with real number coefficients. Since their convergences are fairly complicated, http://blog.yimg.com/2/vyvWdRF7s5_bqHpIzG6otLmp3eg5MBRGEnBi9N0YFnhA34nFqb3srA--/39/o/4oAekeQCUWteJ6Ch0NcAtg.jpg and http://blog.yimg.com/2/vyvWdRF7s5_bqHpIzG6otLmp3eg5MBRGEnBi9N0YFnhA34nFqb3srA--/40/o/52OMY4g6GgniM2CkxG9f1g.jpg have already discussed them in details.
$\left(\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt\right)''-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=\left(-\int_{0}^{\infty}t\sin\left(\frac{t^3}{3}+xt\right)~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=\left(-\int_{0}^{\infty}\frac{t^2\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=\left(-\int_{0}^{\infty}\frac{(t^2+x)\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=\left(-\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~d\left(\frac{t^3}{3}+xt\right)+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=\left(\int_{0}^{\infty}\frac{1}{t}~d\left(cos\left(\frac{t^3}{3}+xt\right)\right)+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=\left(\left[\frac{\cos\left(\frac{t^3}{3}+xt\right)}{t}\right]_{0}^{\infty}-\int_{0}^{\infty}cos\left(\frac{t^3}{3}+xt\right)~d\left(\frac{1}{t}\right)+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=\left(\int_{0}^{\infty}\frac{cos\left(\frac{t^3}{3}+xt\right)}{t^2}~dt+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=-\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt+\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt+x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$
$=0$
