In a mathematical physical problem related to the search of Green's functions in an anisotropic medium, I came across a non-trivial double integral.

Is it possible to evaluate analytically the following double integral?

$$ \varphi(h, A, B) = -\frac{1}{(2\pi)^2} \int_0^{\infty} \int_{0}^\pi \frac{\sin^3\theta \, e^{ihk\cos\theta}}{\cos^4\theta-A\cos^2\theta-B}\, \mathrm{d}\theta \, \mathrm{d}k \, , $$ where $h$, $A$ and $B$ are all positive real numbers. Making the change of variable $q=\cos\theta$, the above integral can conveniently be written as $$ \varphi(h, A, B) = -\frac{1}{(2\pi)^2} \int_0^{\infty} \int_{-1}^1 \frac{(1-q^2) \, e^{ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k \, . $$

Now, by changing the integration range for $q$ we obtain $$ \varphi(h, A, B) = -\frac{1}{2\pi^2} \int_0^{\infty} \int_{0}^1 \frac{(1-q^2) \, \cos(hkq)}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k \, . $$

By making use of Maple, I have noticed that the function $\varphi$ depends solely on $h$ and $B$ and does not depend on $A$. More precisely $$ \varphi (h,A,B) = \varphi(h,B) = \frac{1}{4\pi B h} \, . $$ Is there a way to show that mathematically in a rigorous way? Is that true? Any help / hints / indication would be highly appreciated.

Thank you,


  • 4
    $\begingroup$ Such integral is not convergent in the usual sense. If you are considering the principal value, by partial fraction decomposition, the problem boils down to computing $$ \text{PV}\int_{0}^{+\infty}\int_{-1}^{1}\frac{e^{iqk}}{q^2+C}\,dq\,dk = \text{PV}\int_{-1}^{1}\frac{dq}{q(q^2+C)}=0.$$ $\endgroup$ – Jack D'Aurizio May 16 '17 at 18:35
  • 1
    $\begingroup$ @JackD'Aurizio Thank you. Take for example $A=B=1/2$ then the integral is convergent right? $\endgroup$ – Daddy May 16 '17 at 18:44
  • 1
    $\begingroup$ @JackD'Aurizio The integral above cannot be 0 in my opinion. In fact, represents physically the hydrodynamic pair-mobility between two particles in an anisotropic fluid where $B$ is a viscosity. $\endgroup$ – Daddy May 17 '17 at 8:26
  • 1
    $\begingroup$ You argument has a fatal flaw. You are stating that a mathematical object cannot be zero by using a real-world observation, but your interpretation of reality depends on the mathematical framework you are assuming. What is the physical meaning of $$\int_{0}^{+\infty}e^{iqk}\,dk$$ for instance? Mathematically, that is not a convergent integral in the usual sense. $\endgroup$ – Jack D'Aurizio May 17 '17 at 9:39
  • 2
    $\begingroup$ And the Fourier transform of a function that does not belong to $L^1$ has to be dealt with extreme care. If we cannot apply Fubini's theorem, it means that something strange is happening there, and the issue is not in the real world interpretation, the issue is in the mathematical rigor. What is the Fourier transform (1D, 2D or 3D does not matter) of a function that constantly equals $1$, for instance? Your integrals miss a $\text{PV}$ to actually make sense. $\endgroup$ – Jack D'Aurizio May 18 '17 at 17:46

As pointed out by various comments, the integral is not well-defined in the usual sense if the denominator has zero at some $q \in (-1, 1)$. Instead, we may consider the following regularized version

$$ \tilde{\varphi}(h,A,B) := -\frac{1}{4\pi^2} \lim_{\epsilon \to 0^+} \int_{0}^{\infty} \left( \operatorname{PV} \int_{-1}^{1} \frac{1-q^2}{q^4 - Aq^2 - B} e^{ihkq} \, dq \right) e^{-\epsilon k} \, dk. \tag{1} $$

We write $\alpha = \frac{1}{2}(\sqrt{A+4B} + A)$ and $\beta = \frac{1}{2}(\sqrt{A+4B} - A)$. Then

$$ \frac{1-q^2}{q^4 - Aq^2 - B} = \frac{1-q^2}{(q^2 + \beta)(q^2 - \alpha)}. $$

If $\alpha \geq 1$, then it has no singularity on $[-1 ,1]$ (upon resolving the possible singularity at $q = \pm 1$ when $\alpha = 1$). Otherwise, poles at $p = \pm \sqrt{\alpha}$ should be taken into consideration. So we divide into cases.

Case I. Assume first that $\alpha \geq 1$, which turns the inner principal value into a genuine Lebesgue integral. Under this assumption, we have the uniform estimate

$$ \forall q \in (-1,1), \qquad \left| \frac{1-q^2}{q^4 - Aq^2 - B} \right| =\frac{1-q^2}{(q^2 + \beta)(\alpha - q^2)} \leq \frac{1}{\beta}, \tag{2}$$

hence Fubini's theorem is applicable and we have

\begin{align*} \tilde{\varphi}(h,A,B) &= -\frac{1}{4\pi^2} \lim_{\epsilon \to 0^+} \int_{-1}^{1} \int_{0}^{\infty} \frac{1-q^2}{q^4 - Aq^2 - B} e^{ihkq}e^{-\epsilon k} \, dkdq \\ &= -\frac{1}{4\pi^2} \lim_{\epsilon \to 0^+} \int_{-1}^{1} \frac{1-q^2}{q^4 - Aq^2 - B} \cdot \frac{1}{\epsilon - ihq} \, dq. \end{align*}

Now the inner integral is ready to be computed. Indeed, use symmetry to write

$$ = -\frac{1}{4\pi^2} \lim_{\epsilon \to 0^+} \int_{-1}^{1} \frac{1-q^2}{q^4 - Aq^2 - B} \cdot \frac{\epsilon}{\epsilon^2 + h^2q^2} \, dq. $$

By the standard approximation-to-the-identity argument, we easily check that this limit converges to

$$ = -\frac{1}{4\pi^2} \cdot \frac{\pi}{h} \left( \left. \frac{1-q^2}{q^4 - Aq^2 - B}\right|_{q=0} \right) = \frac{1}{4\pi Bh}. $$

Alternatively, utilize the substitution $hq = \epsilon t$ and notice that the integrand of

$$ \tilde{\varphi}(h,A,B) = -\frac{1}{4\pi^2} \lim_{\epsilon \to 0^+} \int_{-h/\epsilon}^{h/\epsilon} \frac{1-(\epsilon t/h)^2}{(\epsilon t/h)^4 - A(\epsilon t/h)^2 - B} \cdot \frac{1}{h(1+t^2)} \, dt. $$

is dominated by $\frac{1}{\beta h(1+t^2)}$ in view of $\text{(2)}$ again. So we can invoke the dominated convergence theorem to obtain the same answer.

Case II. Assume next that $\alpha < 1$ so that the inner principal value cannot be reduced to ordinary integral. In order to resolve this case, we extract out the contribution of poles at $p = \pm \sqrt{\alpha}$. First, note that

$$ \underset{q = \pm \sqrt{\alpha}}{\operatorname{Res}} \, \frac{1-q^2}{q^4 - Aq^2 - B} = \pm \frac{1-\alpha}{2\sqrt{\alpha}(2\alpha - A)} =: \pm C. $$

Now choose $\delta > 0$ sufficiently small so that $0 < \alpha - \delta < \alpha + \delta < 1$ and an even smooth function $\eta \in C_c^{\infty}(\Bbb{R})$ which is $1$ near $0$ and vanishes outside $(-\delta, \delta)$. Then consider the following function

$$ f(q) = \frac{C}{q - \sqrt{\alpha}} \eta(q - \sqrt{\alpha}) - \frac{C}{q + \sqrt{\alpha}} \eta(q + \sqrt{\alpha}). $$

This function is designed to cancel the poles of $(1-q^2)/(q^4 - Aq^2 - B)$ at $q = \pm \sqrt{\alpha}$, so we can write

\begin{align*} &\operatorname{PV} \int_{-1}^{1} \frac{1-q^2}{q^4 - Aq^2 - B} e^{ihkq} \, dq \\ &\hspace{1em} = \int_{-1}^{1} \left( \frac{1-q^2}{q^4 - Aq^2 - B} - f(q) \right) e^{ihkq} \, dq + \operatorname{PV} \int_{-1}^{1} f(q) e^{ihkq} \, dq. \end{align*}

The integrand of the first term now extends to a continuous as function of $q$ on $[-1, 1]$. So it can be resolved by the same idea as in the former case. This gives

\begin{align*} &-\frac{1}{4\pi^2} \lim_{\epsilon \to 0^+} \int_{0}^{\infty} \left( \int_{-1}^{1} \left( \frac{1-q^2}{q^4 - Aq^2 - B} - f(q) \right) e^{ihkq} \, dq \right) e^{-\epsilon k} \, dk \tag{3} \\ &\hspace{4em} = -\frac{1}{4\pi^2} \cdot \frac{\pi}{h} \left( \left. \frac{1-q^2}{q^4 - Aq^2 - B} - f(q) \right|_{q=0} \right) \\ &\hspace{5em} = \frac{1}{4\pi Bh}. \end{align*}

For the second term, we write

\begin{align*} \operatorname{PV} \int_{-1}^{1} f(q) e^{ihkq} \, dq &= C \cdot \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{ihk(q+\sqrt{\alpha})} - e^{ihk(q-\sqrt{\alpha})}}{q} \, \eta(q) \, dq \\ &= C \int_{-\infty}^{\infty} \frac{\cos (hk(q+\sqrt{\alpha})) - \cos(hk(q-\sqrt{\alpha}))}{q} \, \eta(q) \, dq. \end{align*}

By utilizing trigonometric identities, it is easy to check that the integrand is uniformly bounded by $2kh |\eta(q)|$. So we can invoke Fubini's theorem to write

\begin{align*} &\int_{0}^{\infty} \left( \operatorname{PV} \int_{-1}^{1} f(q) e^{ihkq} \, dq \right) e^{-\epsilon k} \, dk \tag{4} \\ &\hspace{2em} = \int_{-\infty}^{\infty} \frac{C}{q}\left( \frac{\epsilon}{\epsilon^2 + h^2(q+\sqrt{\alpha})^2} - \frac{\epsilon}{\epsilon^2 + h^2(q-\sqrt{\alpha})^2} \right)\, \eta(q) \, dq \\ &\hspace{2em} = - \epsilon \int_{-\infty}^{\infty} \frac{2\sqrt{\alpha}C h^2}{\left( \epsilon^2 + h^2(q+\sqrt{\alpha})^2 \right)\left( \epsilon^2 + h^2(q-\sqrt{\alpha})^2 \right)}\, \eta(q) \, dq \\ &\hspace{2em} = \mathcal{O}(\epsilon) \quad \text{as } \epsilon \to 0^+. \end{align*}

By $\text{(3)}$ and $\text{(4)}$, we again have $\tilde{\varphi}(h,A,B) = 1/(4\pi Bh)$.

Combining altogether, we have proved that

$$ \tilde{\varphi}(h,A,B) = \frac{1}{4\pi Bh} $$


  • When $\alpha \geq 1$, the regularization process in $\text{(1)}$ is unnecessary. Indeed, integration by parts shows that

    $$ \int_{-1}^{1} \frac{1-q^2}{q^4 - Aq^2 - B} e^{ihkq} \, dq = \begin{cases} \mathcal{O}(k^{-2}), & \alpha > 1 \\ \\ -\dfrac{2\sin(hk)}{(1+\beta)hk} + \mathcal{O}(k^{-2}), & \alpha = 1. \end{cases} $$

    Consequently, $\varphi(h, A, B)$ is well-defined for $\alpha > 1$. Then the regularized version reduces to the original version and hence $$\varphi(h,A,B) = \tilde{\varphi}(h,A,B) = \frac{1}{4\pi Bh}. $$

  • On the other hand, the decay speed is worse enough that the exponential regularization in $\text{(1)}$ is essential for $\alpha < 1$. Indeed, a similar comment applies to $\text{(3)}$, showing that

    $$ \int_{-1}^{1} \left( \frac{1-q^2}{q^4 - Aq^2 - B} - f(q) \right) e^{ihkq} \, dq = \mathcal{O}(k^{-2}). $$

    However, the remaining part satisfies

    $$ \operatorname{PV} \int_{-1}^{1} f(q) e^{ihkq} \, dq = -2C \sin (hk\sqrt{\alpha}) \int_{-\infty}^{\infty} \frac{\sin q}{q} \, \eta\left(\frac{q}{hk}\right) \, dq \sim -2C\pi \sin (hk\sqrt{\alpha}) $$

    as $k \to \infty$.


How to motivate the solution

If to convert the integral to the form $$ \varphi(h, A, B) = -\frac{1}{8\pi^2} \int\limits_{-\infty}^{\infty} \int\limits_{-1}^1 \frac{(1-q^2) \, e^{-ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k $$ and then change the order of integration in the integral $$ \varphi(h, A, B) = -\frac{1}{8\pi^2} \int\limits_{-1}^1 \frac{1-q^2}{q^4-Aq^2-B} \left(\int\limits_{-\infty}^{\infty}e^{-2\pi i k\, \dfrac{qh}{2\pi}} \, \mathrm{d}k \right) \, \mathrm{d}q,$$ then there is an opportunity to use the representation of the Dirac delta function as a Fourier transform in the form of $$\delta(x) = \int\limits_{-\infty}^{\infty}e^{-2\pi i k x} \, \mathrm{d}k $$ (see also Wolfram MathWorld),

where $$x={qh\over2\pi},$$ so $$\int\limits_{-\infty}^{\infty}e^{-i hkq} \, \mathrm{d}k = \delta\left({qh\over 2\pi}\right) = {2\pi\over|h|}\delta(q)$$ and $$\boxed{\varphi(h, A, B) = {1\over4Bh}}.$$

How to prove the change in the order of integration

Really, the issue integral is improper and exists as a limit $$\varphi(h, A, B) = -\lim_{M\to\infty}\frac{1}{8\pi^2} \int\limits_{-M}^{M} \int\limits_{-1}^1 \frac{(1-q^2) \, e^{-ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k.$$ Note that the inner integral converges for all positive values of parameters $A$ and $B.$ The external integral in finite limits has no singularities and therefore converges for all possible values of $M,$ so $$\frac{1}{8\pi^2} \int\limits_{-M}^{M} \int\limits_{-1}^1 \frac{(1-q^2) \, e^{-ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k$$ $$ = -\frac{1}{8\pi^2} \int\limits_{-1}^1 \frac{1-q^2}{q^4-Aq^2-B} \left(\int\limits_{-M}^{M}e^{-2\pi i k\, \dfrac{qh}{2\pi}} \, \mathrm{d}k \right) \, \mathrm{d}q.$$

Now the passage to the limit is possible, since as the value of $M$ increases, there are no singularities in the vicinity of the maximum of the function. The properties of the function corresponding to the value of the inner double integral approach, to within a constant, the properties of the delta function (an intense burst with the constant area) and eventually lead to a formally obtained answer. In this case, the possibility of passage to the limit shows that getting integral converges, and this proves the correctness of the change in the order of integration.

  • $\begingroup$ you should motivate how the change of order of integration is valid, otherwise the answer is not that useful $\endgroup$ – tired May 21 '17 at 11:17
  • $\begingroup$ @tired Fixed, thanks $\endgroup$ – Yuri Negometyanov May 21 '17 at 17:43

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