Compute integral $\int^\infty_{-\infty}\frac{1}{\big(\frac89x^2+\frac23\big)\sqrt{\frac{x^2}{3}+1}}~dx$

Question

Problem: I am trying to show that $$\frac1{3\pi}\int^\infty_{-\infty}\frac{dx}{\left(\frac89x^2+\frac23\right)\sqrt{\frac{x^2}{3}+1}}=\frac13$$ This identity is link to the integral of the Airy function on $(0,\infty)$. I have made many attempts (substitution, contour integration ) to show that but no success so far (the integrand has branch singularities that makes integration ab little harder). If someone has a good idea or a trick to find this integral I will appreciate it. Thank you!

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Background: The Airy function can be expressed as a improper Riemann integral $$\operatorname{Ai}(s)=\frac{1}{2\pi}\int^\infty_{-\infty}e^{i\big(x^3/3 + sx\big)}\,dx,\qquad s\in\mathbb{R}$$

It is well known that $\operatorname{Ai}\in L_1(0,\infty)$ and that $$\int_{(0,\infty)}\operatorname{Ai}(s)\,ds =\frac13$$ I am aware some methods to obtain this integral that rely on Laplace's transform and distribution theory (here for example), and also by the relation between the Airy function $\operatorname{Ai}$ and the modified Bessel function of the second kind. The former has the drawback that the validity of change of order of integration is harder to justify; latter is is more to my liking since one can use Fubini-Tonelli's theorem to justify change of order of integration, however it requires knowledge of integral representation of Bessel functions (not a trivial feat).

I am trying to obtain the value of the desired integral by simple methods of Lebesgue integration and Complex Analysis using contour deformation that gives an integrand suitable for Fubini's theorem:

The change of variables $x=s^{1/2}u$ gives $$\operatorname{Ai}(s)=\frac{s^{1/2}}{2\pi}\int^\infty_{-\infty}e^{-s^{3/2}i\big(\frac{u^3}{3}+u\big)}\,du$$ Let $F(z)=\frac{z^3}{3}+z$, $z\in\mathbb{C}$. A simple calculation gives \begin{align} \mathfrak{R}(F)(z)&=x^2y-\frac{y^3}{3}+y\\ \mathfrak{I}(F)(z)&=-\frac{x^3}{3}+xy^2-x \end{align} By deforming the Contour of integration as in method of steepest descent and some estimates, we obtain that \begin{align} \operatorname{Ai}(s)=\frac{s^{1/2}}{2\pi}\int_\gamma e^{-s^{3/2}\mathfrak{R}(F(z))}\,dz=\frac{s^{1/2}}{2\pi}\int^\infty_{-\infty} e^{-s^{3/2}\big(\frac89x^2+\frac23\big)\big(\frac{x^2}{3}+1\big)^{1/2}}\,dx\tag{1} \end{align} where $\gamma$ is the upper branch of the hyperbola $y^2-\frac{x^2}{3}=1$ and where $\mathfrak{I}(F(z))=0$. The integrand in (1) is suitable for Fubini's theorem which gives $$\int^\infty_0\operatorname{Ai}(s)\,ds=\frac{1}{2\pi}\int^\infty_0 s^{1/2}\int_\gamma e^{-s^{3/2}\mathfrak{R}(F(z))}\,dz\,ds=\frac{1}{3\pi}\int^\infty_{-\infty}\frac{dx}{\big(\frac89x^2+\frac23\big)\big(\frac{x^2}{3}+1\big)^{1/2}}$$

Clearly the function $\phi(x)=\frac{1}{3\pi\big(\frac89x^2+\frac23\big)\big(\frac{x^2}{3}+1\big)^{1/2}}$ is Lebesgue integrable and in hindsight $\int_{\mathbb{R}}\phi=\frac13$. I checked my calculation and verified, numerically, that (1) $\phi$ is the correct expression obtained from the method described above, and (2) that indeed it integrates to $1/3$.

Answer 1

Using the substitution $x = \sqrt{3}\sinh t$ lets us rewrite the integrand as

$$\frac{\sqrt{3}}{2\pi}\int_{-\infty}^\infty\frac{dt}{4\sinh^2t+1} = \frac{\sqrt{3}}{2\pi}\int_{-\infty}^\infty\frac{dt}{3\sinh^2t+\cosh^2t} = \frac{1}{2\pi}\int_{-\infty}^\infty\frac{\sqrt{3}\operatorname{sech}^2 t\:dt}{\left(\sqrt{3}\tanh t\right)^2+1}$$

after dividing the numerator and denominator by $\cosh^2 t$. We now have a direct antiderivative:

$$\frac{1}{2\pi}\tan^{-1}\left(\sqrt{3}\tanh t\right)\Bigr|_{-\infty}^\infty = \frac{1}{2\pi}\left(\frac{\pi}{3}+\frac{\pi}{3}\right) = \boxed{\frac{1}{3}}$$

Answer 2

This is a rational integral, hence there is an explicit primitive.

You can find it by giving a rational parametrization of the rational curve $x^2/3+1=y^2$. This can be done fixing a point on the curve, say $(x=0,y=1)$ and considering all lines through that point, i.e. $y=tx+1$, and intersecting again with the curve, $$x^2/3+1=(tx+1)^2\Rightarrow x=\frac {6t}{1-3t^2},\ y=\frac{1+3t^2}{1-3t^2}.$$ Substitution in the integrand gives $$ \frac 1{(\frac 89x^2+\frac 23)y}dx = \frac{9 - 27 t^2}{1 + 42 t^2 + 9 t^4}dt $$ which (e.g. by expanding in partial fractions) has an explicit primitive: $$ \frac{3}{2} \arctan\left(\frac{6 t}{3 t^2+1}\right). $$

Finally, we are interested in the domain of integration $-\infty\leq x\leq +\infty$ and $y\geq 1$, which corresponds to $-\frac1{\sqrt 3}\leq t\leq \frac 1{\sqrt 3}$, yielding the result.

Answer 3

Making the problem more general $(a\neq b)$ $$I(a,b)=\int_{-\infty}^{+\infty} \frac{dx}{\left(x^2+a^2\right) \sqrt{x^2+b^2}}$$

Using, as @Ninad Munshi already made, $x=b\sinh(t)$ $$J=\int \frac{dx}{\left(x^2+a^2\right) \sqrt{x^2+b^2}}=\int \frac{dt}{a^2+b^2 \sinh ^2(t)}$$ $$t=\tanh ^{-1}(u) \quad \implies \quad J=\int \frac{du}{a^2+\left(b^2-a^2\right)u^2}=\frac{\tanh ^{-1}\left(\frac{\sqrt{a^2-b^2}}{a}u\right)}{a \sqrt{a^2-b^2}}$$ $$J=\frac{\coth ^{-1}\left(\frac{a \sqrt{b^2+x^2}}{x \sqrt{a^2-b^2}}\right)}{a \sqrt{a^2-b^2}}$$ making by the end $$\large\color{red}{I(a,b)=\frac{2}{a \sqrt{a^2-b^2}}\coth ^{-1}\left(\frac{a}{\sqrt{a^2-b^2}}\right)}$$

Your problem is $$\frac{3 \sqrt{3}}{8 \pi }\,\,I\left(\frac{\sqrt{3}}{2},\sqrt{3}\right)=\frac 13$$

Edit

For the case of $$K(a,b,c,d)=\int_{-\infty}^{+\infty}\frac{dx}{(x-a) (x-b) \sqrt{(x-c) (x-d)}}$$ where $(a,b,c,d)$ are complex $$K(a,b,c,d)=\frac 4 {b-a}\left(\frac{\tan ^{-1}\left(\frac{\sqrt{c-a}}{\sqrt{a-d}}\right)}{\sqrt{c-a} \sqrt{a-d}}+\frac{\tan ^{-1}\left(\frac{\sqrt{b-c}}{\sqrt{d-b}}\right)}{\sqrt{b-c} \sqrt{d-b}}\right)$$

Answer 4

Note this integral has $\color{red}{\text{anti-derivative}}$, and we can solve it by a simple trig-sub.

Let $x=\sqrt3\tan\theta$,

$$I=\int\frac{3\sqrt3\cos\theta}{8\sin^2\theta+2\cos^2\theta}d\theta=\int\frac{3\sqrt3\cos\theta}{2+6\sin^2\theta}d\theta$$

Let $u=\sqrt3\sin\theta$

$$I=\frac{3}2\int\frac{1}{1+u^2}du=\frac{3}2\arctan(u)=\boxed{\frac{3}2\arctan\left(\frac{\sqrt3~x}{\sqrt{x^2+3}}\right)+C}$$