Leading order asymptotic behaviour of the integral $\int^1_0 \cos(xt^3)\tan(t)dt$

Question

I'm trying to get the leading order asymptotic behaviour of the integral:

$$\int^1_0 \cos(xt^3)\tan(t)dt$$

I'm trying to use the Generalised Fourier Integrals and the Stationary Phase Method, but I can't understand how to start this.

THIS IS WHAT I HAVE TRIED:

We want to get: $$\Re \int^1_0e^{ixt^3}\tan(t)dt$$

$\phi(t)=t^3$ has a stationary point at $t=0$ so the main contribution is around that point and due to the small angle approximation $\tan(t) \approx t$.

$$\Re\int^1_0e^{ixt^3}tdt$$

I tried some substitutions but I can't find anything useful after that.

Answer 1

I'll show a quick way to also include the oscillatory nature of the asymptotic expression in addition to the solution already presented by @Svyatoslav. Having

$$ I = \int_0^1 \exp(ixt^3)\tan(t)dt $$

the minimum of $p(t)=t^3$ occurs at the lower limit $a=0$. Using the notation here.

Taylor expanding around $a=0$ gives that $\tan(t)=t+\frac{t^3}{3}+O(t^5)$. From the link, we identify the parameters $P=1,\mu=3,Q=1,\lambda=2$. Then, to leading term when only considering the point $t=0$, the asymptotic expansion is given by

$$ I(x) \sim \exp(\lambda \pi i/(2\mu))\frac{Q}{\mu}\Gamma\left(\frac{\lambda}{\mu} \right) \frac{\exp(ixp(a))}{(Px)^{\lambda/\mu}} $$

with real part

$$\Re(I) \sim \Gamma\left(\frac{2}{3} \right) \frac{1}{6x^{2/3}}$$

In the case of the upper limit being finite (here $b=1$), an additional term is introduced. Also only considering the leading term (se the mentioned link for the general expression), the full expression is

$$ I(x) \sim \exp(\lambda \pi i/(2\mu))\frac{Q}{\mu}\Gamma\left(\frac{\lambda}{\mu} \right) \frac{\exp(ixp(a))}{(Px)^{\lambda/\mu}} - \exp(ixp(1))P_0(1)\left(\frac{i}{x} \right) $$

with $P_0(t) = \frac{\tan(t)}{3t^2}\Rightarrow P_0(1) = \frac{\tan(1)}{3}$. Now, taking the real part of this gives that

$$ \Re(I) \sim \Gamma\left(\frac{2}{3} \right) \frac{1}{6x^{2/3}} + \frac{\tan(1)}{3} \frac{\sin(x)}{x} $$

This is seen in the figure below, where the legend one term denotes the asymptotic expansion stemming from $a=0$ and two term denotes when also considering the effect of $b=1$.

Answer 2

Let's consider a bit more general case $$I(x,a)=\int_0^1\cos(xt^a)\tan tdt=\Re\int_0^1e^{ixt^a}\tan tdt\overset{xt^a=z}{=}\frac{1}{ax^{1/a}}\Re\int_0^xe^{iz}\tan\left(\frac{z}{x}\right)^\frac1az^{\frac1a-1}dz$$ Decomposing $\tan$ into the series $$=\frac{1}{ax^{1/a}}\Re\int_0^xe^{iz}\left(\big(\frac{z}{x}\big)^{1/a}+\frac{1}{3}\big(\frac{z}{x}\big)^{3/a}+...\right)z^{\frac 1a-1}dz\tag{1}$$ Let's consider the first term: $$I_1=\frac{1}{ax^{1/a}}\Re\int_0^xe^{iz}\Big(\frac{z}{x}\Big)^{1/a}z^{\frac 1a-1}dz=\frac{1}{ax^{2/a}}\Re\int_0^xe^{iz}z^{\frac2a-1}dz\tag{2}$$ Next, we integrate along a quarter-circle in the complex plane: $0\to x\to ix\to i0$, adding the arch of radius $x$. We do not have poles inside our contour; therefore, $$\oint_C=0\,\Rightarrow\,I_1+\frac{1}{ax^{2/a}}\Re\,e^{\frac{\pi i}{a}}\int_x^0e^{-z}z^{\frac2a-1}dz+I_{C_x}=0$$ where $I_{C_x}$ is the integral along the arch of the radius $x$. Extending integration of the second term to $\infty$ (and, therefore, dropping exponentially small corrections), $$I_1\sim\frac{\cos\frac{\pi}{a}\Gamma\left(\frac2a\right)}{ax^{2/a}}-I_{C_x}$$ To estimate $I_{C_x}$, we use the Jordan' lemma. Using $\sin\phi>\frac2\pi\phi$ for $\phi\in[0;\frac\pi2]$, $$|I_{C_x}|=\frac{1}{ax^{2/a}}\left|\int_0^{\pi/2}e^{ix\cos\phi-x\sin\phi}x^\frac2ae^{i\phi\frac2a}id\phi\right|<\frac{1}{a}\int_0^{\pi/2}e^{-\frac2\pi x\phi}d\phi<\frac{\pi}{2ax}$$ We see that $|I_{C_x}|\ll I_1$ at $x\to\infty$ and can be dropped , if $x^\frac2a\ll x$, or $\,\boxed{a>2}$.

In the same way we can evaluate next terms of $\tan$ decomposition and get, for example, $$I(x,a)=\int_0^1\cos(xt^a)\tan tdt\sim\frac{\cos\frac{\pi}{a}\Gamma\left(\frac2a\right)}{a\,x^{2/a}}+\frac{\cos\frac{2\pi}{a}\Gamma\left(\frac4a\right)}{3a\,x^{4/a}},\,\,a>4\tag{3}$$ For $a=3$ we are only allowed to keep the first term: $$\int^1_0 \cos(xt^3)\tan(t)dt\sim\frac{\cos\frac{\pi}{3}\Gamma\left(\frac23\right)}{3x^{2/3}}=\frac16\Gamma\big(\frac23\big)\,x^{-\frac23}\tag{4}$$

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$\bf{Explicite \,form \,of\, full \,asymptotic}$

In his post @AxelT evaluated first oscillating asymptotic term, which gives very good approximation even at $x\sim1$. It would be also interesting to get full asymptotic.

Using integration in the complex plane, we got above $$I(x,a)=\frac{1}{ax^{1/a}}\Re\int_0^xe^{iz}\tan\left(\frac{z}{x}\right)^\frac1az^{\frac1a-1}dz=\frac{1}{ax^{1/a}}\Re \,e^\frac{\pi i}{2a}\int_0^xe^{-z}\tan\left(\frac{z^\frac1a}{x^\frac1a}e^\frac{\pi i}{2a}\right)z^{\frac1a-1}dz$$ $$-\,I_{C_x}\tag{5}$$ where $I_{C_x}$ is the integral along the arch of the radius $x$: $$I_{C_x}=\frac{1}{ax^{1/a}}\Re\int_0^{\pi/2}e^{ixe^{i\phi}}e^{\frac{i\phi}{a}}\tan\big(e^{\frac{i\phi}{a}}\big)x^\frac1ae^{i\phi}id\phi\overset{s=e^{i\phi}}{=}\frac1a\int_1^{e^{\frac{\pi i}{2}}}e^{ixs}\tan(s^\frac1a)s^\frac1ads\tag{6}$$ Integrating (6) several times by part and dropping exponentially small terms ($\sim e^{-\frac{\pi x}{2}}$) $$I_{C_x}\sim\frac{1}{a}\Re\left(-\frac{e^{ix}}{ix}f^{(0)}(s,a)|_{s=1}+\frac{e^{ix}}{(ix)^2}f^{(1)}(s,a)|_{s=1}-\frac{e^{ix}}{(ix)^3}f^{(2)}(s,a)|_{s=1}-...\right)$$ where we denoted $$f^{(k)}(s,a)|_{s=1}:=\frac{\partial^k}{\partial s^k}s^\frac1a\tan(s^\frac1a)\Big|_{s=1}$$ Rearranging $$-I_{C_x}\sim\frac{\sin x}{ax}\left(f^{(0)}(s,a)-\frac{f^{(2)}(s,a)}{x^2}+...\right)\bigg|_{s=1}+\frac{\cos x}{ax^2}\left(f^{(1)}(s,a)-\frac{f^{(3)}(s,a)}{x^2}+...\right)\bigg|_{s=1}$$ $$=\frac1a\sum_{n=1}^\infty\frac{\sin\big(\frac{\pi(n-1)}{2}+x\big)}{x^n}\left(s^\frac1a\tan s^\frac1a\right)^{(n-1)}\,\bigg|_{s=1}\tag{7}$$ Using the series of $\tan$ $$\tan t=\sum_{n=1}^\infty(-1)^{n-1}\frac{4^n(4^n-1)}{(2n)!}B_{2n}t^{2n-1}$$ and integrating term by term (extending integration to $\infty$) $$\frac{1}{ax^{1/a}}\Re \,e^\frac{\pi i}{2a}\int_0^xe^{-z}\tan\left(\frac{z^\frac1a}{x^\frac1a}e^\frac{\pi i}{2a}\right)z^{\frac1a-1}dz\sim\frac1a\sum_{n=1}^\infty\frac{(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}\frac{\cos\frac{\pi n}{a}\Gamma\big(\frac{2n}{a}\big)}{x^\frac{2n}{a}}\tag{8}$$ Putting (7) and (8) into (5), we get the desired asymptotics: $$\color{blue}{I(x,a)\sim\frac1a\sum_{n=1}^\infty\frac{(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}\frac{\cos\frac{\pi n}{a}\Gamma\big(\frac{2n}{a}\big)}{x^\frac{2n}{a}}}$$ $$\color{blue}{+\,\frac1a\sum_{n=1}^\infty\frac{\sin\big(\frac{\pi(n-1)}{2}+x\big)}{x^n}\left(s^\frac1a\tan s^\frac1a\right)^{(n-1)}\,\bigg|_{s=1}}$$ Here we have the only limitation $\color{blue}{a>0}$. If we want to take several first terms, the solution - the sequence of the terms we have to keep - depends, of course, on $a$. For example, for $a=3$ we get the series $$I(x,3)\sim\frac{\cos\frac\pi3\Gamma\big(\frac23\big)}{3x^\frac23}+\frac{\sin x\tan1}{3x}+\frac{\cos\frac{2\pi}{3}\Gamma\big(\frac43\big)}{9x^\frac43}+\frac{2\cos\pi\,\Gamma(2)}{45x^2}+\frac{\cos x}{9x^2}\left(\tan1+\frac{1}{\cos^21}\right)$$ $$+\,O\left(\frac1{x^\frac83}\right)$$

Answer 3

Using the series expansion of $\tan(t)$, we have $$\sum_{n=0}^\infty \frac{(-1)^n 4^{n+1} \left(4^{n+1}-1\right) B_{2 n+2}}{\Gamma (2n+3)}\int_0^1 \cos(xt^3)\,t^{2n+1}\,dx$$ The antiderivative exists and $$\int_0^1 \cos(xt^3)\,t^{2n+1}\,dx=\frac{\, _1F_2\left(\frac{n}{3}+\frac{1}{3};\frac{1}{2},\frac{n}{3}+\frac {4}{3};-\frac{x^2}{4}\right)}{2 (n+1)}$$ which is asymptotically $$\sqrt{\pi } \frac{2^{\frac{2 n-1}{3}} \Gamma \left(\frac{n+1}{3}\right)}{3 \Gamma \left(\frac{3-2n}{6}\right)}x^{-\frac{2}{3} (n+1)} $$ So, for large values of $x$ $$\frac{\Gamma \left(\frac{2}{3}\right)}{6 x^{2/3}}\left(1+\frac{2 \sqrt[3]{2} \sqrt{\pi }}{3 x^{2/3} \Gamma \left(-\frac{1}{6}\right)}\right)+O\left(\frac{1}{x^2}\right)$$