How could this double integral be solved?

$$\int_{x=0}^{\infty}\int_{t=-\infty}^{\infty}\frac{x}{at^2+1}\text{exp}\left(itx-\frac{b \sqrt{3a}t^2}{at^2+1}\right)\mathrm{d}t\ \mathrm{d}x\\ \text{with } a,b>0 \text{ and } i^2=-1$$

  • 1
    $\begingroup$ For the complexity of the problem, maybe you could at least post some more work? $\endgroup$ Commented Dec 19, 2021 at 19:21
  • 2
    $\begingroup$ To start with, you can get rid of $a$ and $\sqrt{3}$ by rescaling variables $\endgroup$
    – Sal
    Commented Dec 19, 2021 at 19:34

1 Answer 1


Some details are below.

After the substitution $t \rightarrow t/\sqrt{a}$ and $x\rightarrow x\sqrt{a}$ the integral becomes $$\int_{x=0}^{\infty}\int_{t=-\infty}^{\infty}\frac{x\sqrt{a}}{t^2+1} \, {\rm exp}\left(itx-\frac{b \sqrt{3/a} \,t^2}{t^2+1}\right)\mathrm{d}t\ \mathrm{d}x \\ =\sqrt{a} \, e^{-c} \int_0^\infty {\rm d}x \, x \, e^{-x} \underbrace{\int_{-\infty}^\infty\frac{e^{ix(t-i)+\frac{c}{t^2+1}}}{t^2+1} \, {\rm d}t}_{I(x)}$$ where $c=b\sqrt{3/a}$. Now $I(x)$ can be closed in the upper half plane since the contribution along the arc vanishes. Then this $t$-integral encloses the single essential singularity in the upper half plane at $t=i$. Hence we have $$I(x)=2\pi i \, {\rm Res} \left(\frac{e^{ix(t-i)+\frac{c}{t^2+1}}}{t^2+1}\right)\Bigg|_{t=i} \, . $$ After some lengthy but basic algebra using the series representation of exp, i.e. $$\frac{e^{ix(t-i)+\frac{c}{t^2+1}}}{t^2+1} = \color{#cc00cc}{\sum_{n=0}^\infty \frac{(ix)^n}{n!} (t-i)^n \sum_{m=0}^\infty \frac{c^m}{m!} \frac{1}{\left[ (t+i)(t-i) \right]^{m+1}}}$$ and reexpanding in powers of $(t-i)$, we eventually arrive at $$I(x)=\pi \sum_{m=0}^\infty \sum_{k=0}^m \frac{x^{m-k}}{(m-k)!} \frac{c^m}{m!^2 2^{m}} \frac{(k+m)!}{2^k k!} \, .$$ We can now carry out the $x$-integral and find $$\int_0^\infty {\rm d}x \, x \, e^{-x} I(x) = \pi \sum_{m=0}^\infty \frac{c^m}{m!^2 2^{m}} \color{#0000f0}{\sum_{k=0}^m (m-k+1) \frac{(k+m)!}{2^k k!}} \\ = \pi \sum_{m=0}^\infty \frac{c^m}{m!^22^m} \cdot \frac{2^{m+1}(m+1/2)!}{\sqrt{\pi}} \\ = 2\sqrt{\pi} \color{#f00000}{\sum_{m=0}^\infty \frac{(m+1/2)!}{m!^2} \, c^m} = {\pi} \, e^{c/2} \left[ (c+1)I_0(c/2) + c \, I_1(c/2) \right]$$ where $I_k$ is the k-th modified Bessel-function.

Some more details:

$$\color{#f00000}{\sum_{m=0}^\infty \frac{(m+1/2)!}{m!^2} \, c^m} = \int_0^\infty {\rm d}t \, e^{-t} t^{1/2} \sum_{m=0}^\infty \frac{(ct)^m}{m!^2} = \int_0^\infty {\rm d}t \, e^{-t} t^{1/2} I_0(2\sqrt{ct}) \\ \stackrel{s=2\sqrt{ct}}{=}\int_0^\infty \frac{s^2}{4c^{3/2}} \, e^{-\frac{s^2}{4c}} I_0(s) \, {\rm d}s = \sqrt{c} \, \frac{{\rm d}}{{\rm d}c} \color{#a0a000}{\int_0^\infty e^{-\frac{s^2}{4c}} I_0(s) \, {\rm d}s}=\sqrt{c} \frac{{\rm d}}{{\rm d}c} \color{#00a000}{\sqrt{\pi c} \, e^{c/2} I_0(c/2)} \, .$$ Now $$\color{#a0a000}{\int_0^\infty e^{-\frac{s^2}{4c}} I_0(s) \, {\rm d}s} = \int_0^\infty e^{-\frac{s^2}{4c}} \int_0^\pi \frac{e^{s\cos(t)}}{\pi} \, {\rm d}t \, {\rm d}s \\ = \sqrt{\frac{c}{\pi}} \int_0^\pi {\rm d}t \, e^{c\cos^2(t)} \left[ 1 + {\rm erf}(\sqrt{c}\cos(t))\right] \stackrel{t=\pi-\tau}{=} \sqrt{\frac{c}{\pi}} \int_0^\pi {\rm d}\tau \, e^{c\cos^2(\tau)} \left[ 1 - {\rm erf}(\sqrt{c}\cos(\tau))\right] \\ \stackrel{\text{symmetry}}{=} \sqrt{\frac{c}{\pi}} \int_0^\pi {\rm d}t \, e^{c\cos^2(t)} = \sqrt{\frac{c}{\pi}} \int_0^\pi {\rm d}t \, e^{c/2\,(\cos(2t)+1)} \stackrel{u=2t}{=} \sqrt{\frac{c}{\pi}} \, e^{c/2} \int_0^{2\pi} \frac{{\rm d}u}{2} \, e^{c/2\cos(u)} \\ \stackrel{\text{symmetry}}{=} \sqrt{\pi c} \, e^{c/2} \int_0^\pi \frac{{\rm d}u}{\pi} \, e^{c/2 \cos(u)} = \color{#00a000}{\sqrt{\pi c} \, e^{c/2} I_0(c/2)} \, .$$

Also: $$\color{#0000f0}{\sum_{k=0}^m (m-k+1) \frac{(k+m)!}{2^kk!}} = \int_0^\infty {\rm d}t \, e^{-t} t^m \sum_{k=0}^m (m-k+1) \frac{t^{k}}{2^kk!} \\ =-\int_0^\infty {\rm d}t \, e^{-t} t^{2m+2} \frac{{\rm d}}{{\rm d}t} \sum_{k=0}^m \frac{t^{k-m-1}}{2^kk!} = -\int_0^\infty {\rm d}t \, e^{-t} t^{2m+2} \frac{{\rm d}}{{\rm d}t} \frac{t^{-m-1} e^{t/2} \Gamma(m+1,t/2)}{m!} \\ \stackrel{{\rm P.I.}}{=} \int_0^\infty {\rm d}t \, \frac{e^{-t/2} t^{m}}{m!} \Gamma(m+1,t/2) \{2(m+1)-t\} \\ \stackrel{t=2s}{=} \frac{2^{m+2}}{m!} \int_0^\infty e^{-s} s^m \Gamma(m+1,s) \{m+1-s\} \, {\rm d}s \\ \stackrel{{\rm P.I.}\text{ in }s^{m+1}\text{-term}}{=} \frac{2^{m+2}}{m!} \int_0^\infty e^{-2s} s^{2m+1} \, {\rm d}s \stackrel{s=t/2}{=} \frac{(2m+1)!}{2^m m!} = \frac{2^{m+1} (m+1/2)!}{\sqrt{\pi}} \, .$$

Reexpanding in powers of $(t-i)$:

$$\color{#cc00cc}{\sum_{n=0}^\infty \frac{(ix)^n}{n!} (t-i)^n \sum_{m=0}^\infty \frac{c^m}{m!} \frac{1}{\left[ (t+i)(t-i) \right]^{m+1}}} = \sum_{n,m=0}^\infty \frac{(ix)^nc^m}{n!m!} \frac{(t-i)^{n-m-1}}{(2i)^{m+1}} \left(\frac{t-i}{2i}+1\right)^{-m-1} \\ = \sum_{n,m,k=0}^\infty \frac{(ix)^nc^m}{n!m!} \frac{(t-i)^{n-m-1+k}}{(2i)^{m+1+k}} \underbrace{\binom{-m-1}{k}}_{=(-1)^k \binom{k+m}{k}} \qquad \qquad \left[\text{Binomial Series}\right] \\ = \sum_{l\in\mathbb{Z}} (t-i)^l \sum_{\substack{n,m,k=0 \\ n+k-m-1=l}}^\infty \frac{(ix)^nc^m}{n!m!} \frac{(-1)^k}{(2i)^{m+1+k}} \binom{k+m}{k} \\ = \sum_{l\in\mathbb{Z}} (t-i)^l \sum_{\substack{n,k=0 \\ n+k \geq l+1}}^\infty \frac{(ix)^nc^{n+k-l-1}}{n!(n+k-l-1)!} \frac{(-1)^k}{(2i)^{n+2k-l}} \binom{n+2k-l-1}{k} \\ = \sum_{l\in\mathbb{Z}} (t-i)^l (2i)^l \sum_{\substack{n,k=0 \\ n+k \geq l+1}}^\infty \frac{x^nc^{n+k-l-1}}{n!(n+k-l-1)!} \frac{1}{2^{n+2k}} \binom{n+2k-l-1}{k}$$ and you can now extract the coefficient of $(t-i)^{-1}$ i.e. $$I(x)=\pi \sum_{k=0}^\infty \sum_{n=0}^\infty \frac{x^nc^{n+k}}{n!(n+k)!} \frac{1}{2^{n+2k}} \binom{n+2k}{k} \stackrel{m=k+n}{=} \pi \sum_{k=0}^\infty \sum_{m=k}^\infty \frac{x^{m-k}}{(m-k)!}\frac{c^{m}}{m!^22^m} \frac{(m+k)!}{2^k k!} \\ = \pi \sum_{m=0}^\infty \sum_{k=0}^m \frac{x^{m-k}}{(m-k)!}\frac{c^{m}}{m!^22^m} \frac{(k+m)!}{2^k k!} \, .$$

  • $\begingroup$ I'm not sure, since I'm not familiar with such type of integrals. Maybe someone else is better suited to answer this. Where did it arise? $\endgroup$
    – Diger
    Commented Dec 20, 2021 at 18:31
  • $\begingroup$ Also related: math.stackexchange.com/questions/4338561/… $\endgroup$ Commented Dec 26, 2021 at 5:26
  • $\begingroup$ How do we go from the last step of reexpanding in powers of $(t-i)$ to $I(x)$ in the main text? It is still a quite skillful derivation, and I placed a bounty on the related question math.stackexchange.com/q/4341769/737314 $\endgroup$ Commented Dec 27, 2021 at 19:28
  • $\begingroup$ clever derivation indeed (+1) $\endgroup$
    – G Cab
    Commented Jan 11, 2022 at 0:14

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