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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.

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  • $\begingroup$ One way is to simply check the conditions of the DCT. $\endgroup$
    – user21820
    Jun 28, 2019 at 9:41
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    $\begingroup$ A more concrete way is to use hard bounds on the integrand, so that you can integrate them and then take the limit. $\endgroup$
    – user21820
    Jun 28, 2019 at 9:43
  • $\begingroup$ But in this case I will obtain a bound of the limit and I want to get its value. $\endgroup$
    – popi
    Jun 28, 2019 at 10:07
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    $\begingroup$ @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. $\endgroup$
    – Alex M.
    Jun 28, 2019 at 15:28
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    $\begingroup$ @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. $\endgroup$
    – user21820
    Jun 28, 2019 at 16:07

2 Answers 2

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

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  • $\begingroup$ 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? $\endgroup$
    – popi
    Jun 30, 2019 at 22:35
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    $\begingroup$ 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. $\endgroup$
    – Maxim
    Jun 30, 2019 at 23:01
  • $\begingroup$ 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. $\endgroup$
    – popi
    Jul 1, 2019 at 7:39
  • $\begingroup$ Do yuo think that could be usefull the Stepest descent method to obtain a convergent expansion of this integral? $\endgroup$
    – popi
    Jul 2, 2019 at 8:31
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    $\begingroup$ One may note that the last integral is related to the Fransen–Robinson constant, see here. $\endgroup$
    – p4sch
    Jul 4, 2019 at 2:59
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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.

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  • $\begingroup$ 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? $\endgroup$
    – popi
    Jul 4, 2019 at 14:38

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