I'm trying to calculate the following integral (using special functions and/or contour integration):

$$I(x)=\int_{-\infty}^{\infty}e^{ix t}t\,e^{-i\sqrt{t^2+a^2}} \,dt$$

Here, $x$ is a real number/variable, $a$ is a real number, and $i$ is the imaginary unit. I'm only really interested in the $x\rightarrow \infty$ ($x$ is a real number) limiting behavior of this function, but it would be especially nice if I could get some sort of closed form expression for this integral.

If you also know of some particular method I could use to calculate this integral, please tell me. In the form that I currently have it in, I am compelled to think that some sort of an asymptotic expansion may work. I suppose I could also change the bounds of integration, write $t=\exp (\ln(t))$, and try the method of steepest descents.

  • $\begingroup$ is $z$ assumed to be a complex number? $\endgroup$ – Dr. Sonnhard Graubner Feb 1 '18 at 13:09
  • $\begingroup$ @Dr.SonnhardGraubner No, it isn't, sorry. $\endgroup$ – Arturo don Juan Feb 1 '18 at 13:54
  • $\begingroup$ @Dr.SonnhardGraubner Yes, it is. $\endgroup$ – Arturo don Juan Feb 1 '18 at 14:03
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    $\begingroup$ Don't you see that the integral doesn't converge? $\endgroup$ – Sigma Algebra Feb 1 '18 at 14:12
  1. OP's integrand in his integral $$I(x)~:=~\int_C\! \mathrm{d}w~we^{i\{xw -\sqrt{w^2+a^2}\}},$$ $$ w~\equiv~u+iv, \qquad u,v~\in~\mathbb{R}, \qquad x~>~1, \qquad a~\geq~ 0,\tag{1}$$ is not ${\cal L}^1$-integrable if we chose the integration contour $C$ to be the real $u$-axis in the complex $w$-plane. To repair this, we slightly deform the integration contour $C$ into the upper half-plane $$u\quad\mapsto\quad w~=~u+i\underbrace{\epsilon|u|}_{=v}, \qquad u~\in~\mathbb{R}. \tag{2} $$ In this answer we will only consider OP's integral (1) with the integration contour (2). Here $\epsilon>0$ is a sufficiently small regularization parameter, which the integral (1) does not depend on.

  2. Next issue is that the square root in the exponential (1) is double-valued. We pick the convention that the imaginary part of a square root $\sqrt{z}$ has the same sign as its argument $z$. In other words, we have a branch cut at the negative ${\rm Re}(z)$-axis. This translates into a branch cut at the imaginary $v$-axis in the complex $w$-plane for $|v|\geq a$.

  3. We next deform the integration contour to surround the upper branch cut in the complex $w$-plane. The result is $$\begin{align} I(x)&~~~~=~\int^{v=a}_{v=\infty} \mathrm{d}(iv)~(iv)e^{i\{x(iv)+i\sqrt{v^2-a^2}\}} +\int_{v=a}^{v=\infty} \mathrm{d}(iv)~(iv)e^{i\{x(iv)-i\sqrt{v^2-a^2}\}} \cr &~~~~=~-2\int_a^{\infty}\! \mathrm{d}v~ve^{-xv}\sinh\sqrt{v^2-a^2} \cr &\stackrel{v=t/x+a}{=}~ -\frac{2e^{-xa}}{x}\int_{\mathbb{R}_+}\! \mathrm{d}t~e^{-t}\left(\frac{t}{x}+a\right)\sinh\sqrt{\frac{t}{x}\left(\frac{t}{x}+2a\right)} \cr &~~~~=~\left\{\begin{array}{rcl} -\sqrt{2\pi}e^{-xa}\left(\frac{a}{x}\right)^{3/2} \left\{1+O(x^{-1})\right\} &{\rm for}& a~>~0, \cr \cr -\frac{4}{x^3}\left\{1+O(x^{-1})\right\} &{\rm for}& a~=~0, ,\end{array}\right. \end{align}\tag{3}$$ where we in the last expression gave the leading asymptotic behaviour for $x\to\infty$.

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Naive reasoning: since you're interested in the case $x \to +\infty$ then the exponential may shrink just into



$$\int_{-\infty}^{+\infty} t\ e^{itx}\ dx = -i\sqrt{2\pi}\delta'(x)$$

Being it the Fourier Transform of the function $t$.

$\delta'$ is the derivative of Dirac Delta function.

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  • $\begingroup$ I don't quite understand this reasoning. What do you mean "the exponential may shrink into..."? $\endgroup$ – Arturo don Juan Feb 1 '18 at 14:06
  • $\begingroup$ I mean that since $x\to \infty$, the rest gets neglected. $\endgroup$ – Sigma Algebra Feb 1 '18 at 14:13

The integral is equivalently given as the Fourier transform

$$ I_a(x) = \int_{-\infty}^{+\infty} e^{ixt} t e^{-i\sqrt{t^2+a^2}}dt = \mathcal F[\sqrt{2\pi}te^{-i \sqrt{t^2 + a^2}}](-x) $$

If $a=0$ then $e^{-i \sqrt{t^2 + a^2}}$ simplifies to $e^{-i|t|}$ and the integral is analytically solveable:

$$I_0(x) = \frac{-4x}{(x^2-1)^2}$$

Giving $\lim_{x\to\infty} I_0(x) = 0$. I suspect the same holds for all $a$ but I can't prove it quite yet.

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  • $\begingroup$ Thanks! This is interesting! I'm also interested in the leading order behavior (above $0^{\text{th}}$ order). $\endgroup$ – Arturo don Juan Feb 1 '18 at 14:58

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