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I'm trying to calculate the following limit

\begin{equation}\label{eq}\large\lim_{R\to\infty}\,i\,\int_{-\pi/2}^{\pi/2}\frac{e^{-\alpha(R+i\,w)}e^{t\,e^{R+i\,w}}}{(R+i\,w)^{\beta}}dw\end{equation}

with $\alpha\geq0$, $\beta, t>0$ .

The problem is that (I think, I'm not sure...) I can't change limit by integral.

Then, I have 2 doubs:

  • Can I change limit by integral?

  • If the answer is not, another way to manipulate this limit is welcomed.

Update 1:

Using the generating function of Bell polynomials of first kind $B_n(t)$

$$\large e^{t(e^u-1)}=\sum_{n=0}^\infty\frac{B_n(t)}{n!}u^n$$

reemplacing $u \rightarrow R+i\,w$ we have

$$\lim_{R\to\infty}\,i\,\sum_{n=0}^\infty\frac{e^t\,B_n(t)}{n!}\int_{-\pi/2}^{\pi/2}e^{-\alpha(R+i\,w)}\,(R+i\,w)^{n-\beta}dw=$$

$$\large\lim_{R\to\infty}\sum_{n=0}^\infty\frac{e^t\,B_n(t)}{n!\,\alpha^{n-b-1}}\left[\,\Gamma(n-\beta-1,\alpha(R-i\,\pi/2))-\Gamma(n-\beta-1,\alpha(R+i\,\pi/2))\,\right]$$

Update 2:

Changing the variable $R+i\,w\rightarrow u$ we have

$$\large\lim_{R\to\infty}\,i\,\int_{-\pi/2}^{\pi/2}\frac{e^{-\alpha(R+i\,w)}e^{t\,e^{R+i\,w}}}{(R+i\,w)^{\beta}}dw=\lim_{R\to\infty}\,\int_{R-i\,\pi/2}^{R+i\,\pi/2}\frac{e^{-\alpha\,u}e^{t\,e^u}}{u^{\beta}}du$$

and maybe we can apply complex integration (Cauchy theorem,...) in the last one.

popi
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  • One way is to simply check the conditions of the DCT. – user21820 Jun 28 '19 at 09:41
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    A more concrete way is to use hard bounds on the integrand, so that you can integrate them and then take the limit. – user21820 Jun 28 '19 at 09:43
  • But in this case I will obtain a bound of the limit and I want to get its value. – popi Jun 28 '19 at 10:07
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    @popi: In order to do what you want you should make sure that the integrand has a limit. If it had, its abolute value also would. But notice that the dominant part in the integrand is $\exp [t \exp(R)]$ which tends to $\infty$. The other parts of the integrand that tend to $0$ (such as $\exp (- \alpha R)$) do so much more slowly (polynomially versus exponentially), not fast enough to cancel the part tending to $\infty$. The integrand, therefore, has no limit. The limit of the integral might still exist, though; you just may not interchange the limit and the integral. – Alex M. Jun 28 '19 at 15:28
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    @popi: Not necessarily. I did not attempt to solve your problem, but there are instances where the bounds are asymptotically tight enough that integrating yields bounds that have the same limit. Given Alex's remark, it's worth a try. – user21820 Jun 28 '19 at 16:07
  • @popi: Is $\beta$ an integer, by any chance? This might simplify things a tiny bit. – Alex M. Jun 28 '19 at 18:43
  • @Alex M No, $\beta$ is a complex parameter whose real part is bigger than 0 – popi Jun 28 '19 at 19:06
  • I have just added some advances. – popi Jun 28 '19 at 22:33

2 Answers2

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Take $z = e^{R + i w}$. The integrand becomes $f(z) = e^{t z} z^{-\alpha - 1} \ln^{-\beta} z$, and $$I(R) = \int_{\gamma_1} f(z) \, dz = \int_{\gamma_2} f(z) \, dz.$$ The integrals over the arcs of the left semicircle tend to zero, therefore $$I = \lim_{R \to \infty} I(R) = \int_{\gamma(1)} f(z) \, dz.$$ If $\beta = 1$, then $$I = 2 \pi i e^t + \int_{\gamma(0)} f(z) \, dz.$$

Alternatively, since $I$ can be converted into the Bromwich integral, $$I = 2 \pi i \mathcal L^{-1}[z \mapsto z^{-\alpha - 1} \ln^{-\beta} z](t), \\ I \bigg\rvert_{(\alpha, \beta, t) = (0, 1, 1)} = 2 \pi i \int_0^1 \int_0^\infty \frac {\tau^{u - 1}} {\Gamma(u)} \, du d\tau = 2 \pi i \int_0^\infty \frac {du} {\Gamma(u + 1)}.$$

Maxim
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  • Thanks for the answer, it's very interesting. So, talking about $\int_{\gamma(1)} f(z) , dz$, do you know how to evaluate the integral on the horizontal lines? – popi Jun 30 '19 at 22:35
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    It's unlikely to have a nice closed form. If you take $\alpha < 0$, in which case the integral over $\gamma(0)$ becomes $\int_{-\infty - i 0}^0 + \int_0^{-\infty + i 0}$, you'll get $\ln^2 z + \pi^2$ in the denominator, same as in your other related questions. – Maxim Jun 30 '19 at 23:01
  • This is also my big problem, I got a similar expression (with the horizontal lines on the right of vertical line) but I don't know how compute the integral over the horizontal lines. It is a big wall that I can't cross. – popi Jul 01 '19 at 07:39
  • Do yuo think that could be usefull the Stepest descent method to obtain a convergent expansion of this integral? – popi Jul 02 '19 at 08:31
  • An expansion in which parameter? We have already eliminated $R$. If you're interested in the case $(\alpha, \beta, t) = (0, 1, 1)$, the integral of $1/\Gamma$ already converges extremely fast. – Maxim Jul 02 '19 at 10:56
  • An expansión of type $\sum_{n=0}^{\infty}a_n(\alpha,\beta,t)$ convergent. Steepest descent method is usually used to obtain asymptotic exapnsion but sometimes provides convergent expansions (related with the radius of convergence of the function wich one develop ). What do you think about? – popi Jul 02 '19 at 12:29
  • If $t$ is large, we just need to estimate the integral over a small neighborhood of $z = 1$. Since the non-exponential factors have simple expansions there, we get a simple leading term. Is that useful? You do not specify whether the goal is numerical evaluation, or estimates for large/small values of some of the parameters, or something else. – Maxim Jul 02 '19 at 13:44
  • As the original limit seems to be impossible to calculate, I would like to get an analytical approximation, more specific, a convergent expansion. Later, It could be interesting approximate for large values of parameters(asymptotic), but I think this is easier. – popi Jul 02 '19 at 15:23
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    One may note that the last integral is related to the Fransen–Robinson constant, see here. – p4sch Jul 04 '19 at 02:59
  • Yes, it seems similar. – popi Jul 04 '19 at 08:41
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We have $-\frac{\pi}{2}<w<\frac{\pi}{2}$ and $$ \left|\frac{e^{-\alpha(R+iw)}\exp\left(te^{R+iw}\right)}{(R+iw)^\beta}\right|=\left|\frac{e^{-\alpha R}e^{-i\alpha w}\exp\left(te^R(\cos w+i\sin w)\right)}{(R+iw)^\beta}\right|= $$ $$ =e^{-aR}\exp\left(te^{R}|\cos w|\right)\left|\exp\left(ite^R\sin w\right)\right|\frac{1}{(\sqrt{R^2+w^2})^{\beta}}= $$ $$ =\frac{e^{-aR}\exp\left(te^{R}|\cos w|\right)}{(\sqrt{R^2+w^2})^{\beta}}=\frac{\exp\left(|\cos w|te^{R}-aR\right)}{(R^2+w^2)^{\beta/2}}>>\exp(t|\cos w|e^{R})\textrm{, }R\rightarrow\infty\tag 1 $$ Then also $$ i\int^{\pi/2}_{-\pi/2}f(R,w)dw=i\int^{\pi/2}_{-\pi/2}|f(R,w)|e^{i\theta(R,w)}dw= $$ $$ =-\int^{\pi/2}_{-\pi/2}|f(R,w)|\sin(\theta(R,w))dw+i\int^{\pi/2}_{-\pi/2}|f(R,w)|\cos(\theta(R,w))dw. $$ Assume now we can interchange the limit and integral. Since $\sin(\theta)\geq -1$, for all $\theta\in \textbf{R}$, we get $$ Re\left(\lim_{R\rightarrow\infty}i\int^{\pi/2}_{-\pi/2}f(R,w)dw\right)=Re\left(\int^{\pi/2}_{-\pi/2}\lim_{R\rightarrow\infty}f(R,w)dw\right)\geq $$ $$ \geq\int^{\pi/2}_{-\pi/2}\lim_{R\rightarrow\infty}|f(R,w)|dw=\infty\textrm{, from relation }(1). $$ Hence we can not interchange limit and integral.

Nikos Bagis
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  • Thank you Nikos: is not possible interchange limit and integral as we suspected. Now, we will have to calculate the limit in another way but, how to do this? – popi Jul 04 '19 at 14:38