Finding the value of the following double integral

Question

The following question appeared in the American Mathematical Monthly (AMM), problem 12247, Vol.128, April 2021.

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For positive real constants $a$, $b$, and $c$, prove $$\int_0^{\pi} \int_0^{\infty} \frac{a}{\pi(x^2+a^2)^{3/2}} \frac{x}{\sqrt{x^2+b^2+c^2-2cx\cos \theta}}~dx~d\theta=\frac{1}{\sqrt{(a+b)^2+c^2}}.$$

I tried to use the substitution $x=a\tan\theta$. But it does not help a lot. Please guide. Any answer will be highly appreciated.

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Solutions posted in MSE are based on Fourier transform methods which sometimes are beyond the scope of undergraduate courses. I would like to see if solutions based on more traditional Calculus methods can be found.

Answer 1

The proposed expression can be interpreted as an integral in the superior half-plane in polar coordinates $(\rho,\theta)$. It is transformed below into cartesian coordinates $(x=\rho\cos\theta,y=\rho\sin\theta)$ \begin{align} I&=\frac{a}{\pi}\int_0^{\pi} \int_0^{\infty} \frac{1}{(\rho^2+a^2)^{3/2}} \frac{1}{\sqrt{\rho^2+b^2+c^2-2c\rho\cos \theta}}\,\rho d\rho~d\theta\\ &=\frac{a}{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{(x^2+y^2+a^2)^{3/2}} \frac{1}{\sqrt{(x-c)^2+y^2+b^2}}\,dxdy \end{align} Under this form, it appears as a convolution integral which can be evaluated by a 2D Fourier transform technique. The integral $I$ writes \begin{align} I&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(x,y)g(x-c,y)\,dxdy\\ f(x,y)&=\frac{a}{2\pi}\frac{1}{(x^2+y^2+a^2)^{3/2}}\\ g(x,y)&= \frac{1}{\sqrt{x^2+y^2+b^2}} \end{align} Both functions are radially symmetric, their Fourier transforms can be evaluated using Hankel transforms (with $k=\sqrt{k_x^2+k_y^2}$): \begin{align} \hat{f}(k_x,k_y)&=a\int_0^\infty \frac{J_0(kr)}{(r^2+a^2)^{3/2}}\,rdr\\ &=e^{-ka}\\ \hat{g}(k_x,k_y)&=2\pi\int_0^\infty \frac{J_0(kr)}{(r^2+b^2)^{1/2}}\,rdr\\ &=2\pi \frac{e^{-kb}}{k} \end{align} Then, performing the inverse Fourier transform, \begin{align} I=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(a+b)k}e^{-2i\pi k_xc}\,\frac{dk_xdk_y}{k} \end{align} This integral can be evaluated in polar coordinates: \begin{align} I&=\int_{0}^{\infty} \int_{0}^{2\pi} e^{-(a+b)k}e^{-2i\pi kc\cos\phi}\,d\phi dk\\ &=2\pi\int_{0}^{\infty}e^{-(a+b)k}J_0(2 \pi kc)\,dk\\ &=\frac{1}{\sqrt{(a+b)^2+c^2}} \end{align} as expected

Answer 2

This is to address integrability issues not addressed in @PaulEnta answer to this problem.

As observed, after a change of variables using polar coordinates , the desired integral $I$ may be expressed as the convolution at $(c,0)$ of the functions \begin{align} f_a(x,y)&=\frac{a}{2\pi}\frac{1}{(x^2+y^2+a^2)^{3/2}}\\ g_b(x,y)&= \frac{1}{\sqrt{x^2+y^2+b^2}} \end{align} that is, $I=(f_a*g_b)(c,0)=\int_{\mathbb{R}^2}f_a(x,y)g_b(c-x,-y)\,dxdy$

Connections to well known Fourier transforms:

1. It is known from Fourier analysis (Stein and Weiss for instance) that $f_a$ is related to Poisson kernel in 2D, and that it is the Fourier transform of a $L_1$ function:

Let $\Phi_a(\boldsymbol{x})=e^{-2\pi a|\boldsymbol{x}|}$, which is in $L_1(\mathbb{R}^2)$.

$$ \widehat{\Phi_1}(\boldsymbol{y})=\int_{\mathbb{R}^2} e^{-2\pi i \boldsymbol{x}\cdot\boldsymbol{t}} e^{-2\pi|\boldsymbol{x}|},d\boldsymbol{x} =\frac{1}{2\pi}\frac{1}{(1+|\boldsymbol{y}|^2)^{3/2}} $$ and so, $$ \widehat{\Phi_a}(\boldsymbol{y})=a^{-2}\widehat{\Phi_1}(a^{-1}\boldsymbol{y})=\frac{a}{2\pi(a^2+|\boldsymbol{y}|^2)^{3/2}}=f_a(\boldsymbol{y})$$ Observed that $f_a\in L_1(\mathbb{R}^2)$ as a simple calculation using polar coordinates shows.

2. The function $g_b$ is the Fourier transform of an $L_1$ function:

Let $\Psi_b(\boldsymbol{x})=\frac{e^{-2\pi b|\boldsymbol{x}|}}{2\pi b|x|}$ for $\boldsymbol{x}\neq\boldsymbol{0}$, which is in $L_1(\mathbb{R}^2)$.

$$ \widehat{\Psi_1}(\boldsymbol{y})=\frac{1}{2\pi \sqrt{1+|\boldsymbol{y}|^2}} $$ This particular Fourier transform is computed here. Hence $$ \widehat{\Psi_b}(\boldsymbol{y})=b^{-2}\widehat{\Phi_1}(b^{-1}\boldsymbol{y}) =\frac{1}{2\pi b \sqrt{b^2+|\boldsymbol{y}|^2}}=\frac{1}{2\pi b}g_b(\boldsymbol{y}) $$ Notice however that $g_b\not\in L_1(\mathbb{R}^2)$.

3. By Young's convolution theorem $f_a*g_b\in\mathcal{C}_\infty(\mathbb{R}^2)$. In order to invoke the Fourier inversion theorem, some integrability considerations are in order:

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Lemma: Assume $\Phi,\Psi\in L_1(\mathbb{R}^n)$ and $\widehat{\Phi}\in L_1(\mathbb{R}^n)$. Then $\Phi\cdot\Psi\in L_1(\mathbb{R}^n)$, and \begin{align} \widehat{\Phi\cdot\Psi}=\widehat{\Phi}*\widehat{\Psi}\tag{1}\label{one} \end{align} Proof: The Fourier inversion theorem for $L_1$ implies that $\Phi$ is almost surely equal to function (which we still denote by $\Phi$) in $\mathcal{C}_\infty(\mathbb{R}^n)\cap L_1(\mathbb{R}^n)$ and $$ \Phi(\boldsymbol{x})=\int_{\mathbb{R}^n}\widehat{\Phi}(\boldsymbol{y})e^{2\pi i \boldsymbol{x}\cdot\boldsymbol{y}}\,d\boldsymbol{y} $$

It follows that $\Phi\cdot\Psi\in L_1(\mathbb{R}^n)$ and so,

\begin{align} \widehat{\Phi\cdot\Psi}(\boldsymbol{t})=\int_{\mathbb{R}^n}\Phi(\boldsymbol{x})\Psi(\boldsymbol{x}) e^{-2\pi\boldsymbol{x}\cdot\boldsymbol{t}}\,d\boldsymbol{x}=\int_{\mathbb{R}^n}\Psi(\boldsymbol{x}) e^{-2\pi\boldsymbol{x}\cdot\boldsymbol{t}}\Big(\int_{\mathbb{R}^n}\widehat{\Phi}(\boldsymbol{y})e^{2\pi i \boldsymbol{x}\cdot\boldsymbol{y}}\,d\boldsymbol{y}\Big) \,d\boldsymbol{x} \end{align} Since $(\boldsymbol{x},\boldsymbol{y})\mapsto \Psi(\boldsymbol{x})\widehat{\Phi}(\boldsymbol{y})$ is in $L_1(\mathbb{R}^n\times\mathbb{R}^n)$, we can apply Fubini's theorem to obtain \begin{align} \widehat{\Phi\cdot\Psi}(\boldsymbol{t})= \int_{\mathbb{R}^n}\widehat{\Phi}(\boldsymbol{y}) \Big(\int_{\mathbb{R}^n}\Psi(\boldsymbol{x})e^{-2\pi i \boldsymbol{x}\cdot(\boldsymbol{t}-\boldsymbol{y})}\,d\boldsymbol{x}\Big) \,d\boldsymbol{y}=\int_{\mathbb{R}^n}\widehat{\Phi}(\boldsymbol{y}) \widehat{\Psi}(\boldsymbol{t}-\boldsymbol{y})\,d\boldsymbol{y}=\widehat{\Phi}*\widehat{\Psi}(\boldsymbol{t}). \end{align}

Remark: The equality in \eqref{one} equality among continuous functions (pointwise).

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4. Now that we have taken all integrability considerations out of the way, we can proceed as in @PaulEnta's (our lemma outlines the strategy) \begin{align} f_a*g_b(t_1,t_2)&=2\pi b\widehat{\Phi_a\cdot\Psi_b}(t_1,t_2)=2\pi b\int_{\mathbb{R}^2}\frac{e^{-2\pi|\boldsymbol{x}|(a+b)}}{2\pi b|\boldsymbol{x}|}e^{-2\pi i\boldsymbol{x}\cdot(t_1,t_2)}\,d\boldsymbol{x}\\ &=2\pi(a+b)\widehat{\Psi_{a+b}}(t_1,t_2)=\frac{1}{\sqrt{(a+b)^2+t^2_1 + t^2_2 }} \end{align}

Answer 3

Rewrite the integral in Cartesian coordinates

\begin{align} I = & \frac a\pi\int_0^\infty \int_{-\infty}^{\infty} \frac{1}{(x^2+y^2+a^2)^{3/2}} \frac{dxdy}{\sqrt{(x-c)^2+y^2+ b^2}}\\ \end{align} Substitute $y=\sqrt{(x-c)^2+b^2} \sinh t$

$$I= \frac a\pi\int_0^\infty \int_{-\infty}^{\infty} \frac{1}{[(x\cosh t + c \>\text{sech}t)^2+b^2\sinh^2 t + c^2\tanh^2 t]^{3/2}} dx \>dt $$

Substitute $u= x\cosh t + c \>\text{sech}\>t $ and integrate over $u$ with $\int_{-\infty}^{\infty} \frac{du}{(u^2+p^2)^{3/2} }= \frac2{p^2} $

$$I =\frac {2a}\pi\int_0^\infty \frac{\cosh t}{a^2+(a^2+b^2+c^2)\sinh^2 t + b^2\sinh^4 t}dt\\ $$ Substitute $s=\cosh t$ and then integrate over $s$ to obtain \begin{align} I=\frac {2a}\pi\int_0^\infty \frac{ds}{a^2+(a^2+b^2+c^2)s^2 + b^2s^4} =\frac{1}{\sqrt{(a+b)^2+c^2}} \end{align}