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I have computed the density function from the distribution function and used the definition of characteristic functions. I am stuck at the following integral.$$\varphi_X(t)=\int_{-\infty}^{+\infty} \frac{e^{(1+it)x}}{(1+e^x)^2} dx$$

I am given the following two hints:

  1. $B(x,y)=\int_{0}^{+\infty}\frac{t^{x-1}}{(1+t)^{x+y}}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ for $\text{Re}(x)>0$ and $\text{Re}(y)>0$.

  2. $\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)}$ for $z\in\mathbb{C}$.

A hint on how to proceed with this integral would be much appreciated!

Leon
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    Have you tried anything? – whpowell96 Aug 18 '23 at 18:30
  • The usual step in this case is a substitution such as $u=e^x$ or $y=e^{-x}$ – Henry Aug 18 '23 at 18:53
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    Note the density is an even function, so the characteristic function will also be even. – Thomas Andrews Aug 18 '23 at 18:56
  • Contour integration seems like something that would work. Use a rectangular contour. It should come out to be $\pi t \operatorname{csch}(\pi t)$ – Captain Chicky Aug 18 '23 at 18:58
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    Wait ah, you're doing it the beta function way. Nevermind then the contour integration. Use substitutions to convert your integral into the beta function form, split the beta into gammas, and then apply reflection. – Captain Chicky Aug 18 '23 at 19:01
  • @CaptainChicky Apologies for this is probably elementary, but I’m not sure how to use substitutions to obtain the form needed for the first hint. – Leon Aug 18 '23 at 19:07
  • You may try contour integration and calculus of residues. – Mittens Aug 18 '23 at 19:10
  • @Mittens I believe this integral should be solved without having recourse to contour integration (just using the two provided hints and standard calculus). – Leon Aug 18 '23 at 19:12
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    I'll take a look with the beta method and write an answer hmm – Captain Chicky Aug 18 '23 at 19:16
  • @CaptainChicky Thank you! – Leon Aug 18 '23 at 19:17
  • @Leon: the identity (2) is not standard Calculus (unless you studied at Moscow State University) as it requires some Complex Analysis. In any event, there are many ways to get. the integral. – Mittens Aug 18 '23 at 19:24
  • this got closed after I edited my answer kekw rip average MSE experience ig – Captain Chicky Aug 19 '23 at 19:35

1 Answers1

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There are two ways you can do this.

You can try contour integration, in which you would closely follow the steps for a rectangular contour detailed here, but with higher order poles.

Or, the beta function method you have been hinted for. To do that, let me first establish $a=1+it$ because it's easier to write. The integral becomes $$\int_{-\infty}^{\infty} \frac{e^{ax}}{(1+e^x)^2} dx = \int_{-\infty}^{\infty} \frac{\left(e^{x}\right)^a}{(1+e^x)^2} dx$$ Enforce the substitution $$u=e^x$$ $$\frac{du}{dx}=e^x,\quad \frac{dx}{du}=\frac1{e^x},\quad dx=\frac{du}{e^x}=\frac{du}{u}$$ The bounds transform as follows. Lower one is $e^{-\infty}=0$, and upper one is $e^\infty=\infty$.
(More formally, the differential manipulation can be justified with nonstandard analysis and the directly plugging in infinity should instead be done with a limit, but you get what I mean since this is pretty standard substitution i hope?)

Using this substitution yields $$\int_0^\infty\frac{u^a}{(1+u)^2}\cdot\frac{du}{u} = \int_{0}^{\infty} \frac{u^{a-1}}{(1+u)^2} du = \int_{0}^{\infty} \frac{u^{(1+it)-1}}{(1+u)^2} du$$

Since the beta function is given as $$\operatorname{B}(x,y)=\int_{0}^{\infty}\frac{t^{x-1}}{(1+t)^{x+y}}dt=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$$ we have $$x = 1+it,\quad y=1-it$$

Our integral now becomes $$\int_{0}^{\infty} \frac{u^{(1+it)-1}}{(1+u)^{(1+it)+(1-it)}} du = \text{B}(1+it, 1-it) = \frac{\Gamma(1+it)\,\Gamma(1-it)}{\Gamma(2)}$$

This is just $$\Gamma(1+it)\,\Gamma(1-it)$$

I believe you can take it here on your own. Additional hints below ig.

Use the property that $\Gamma(1+x) = x\,\Gamma(x)$.

Apply it on $\Gamma(1+it)$ and then use your second hint, which is Euler's Reflection Formula. Manipulate it and solve for the Gamma duo, then plug your result in. It should simplify down to a hyperbolic function (this way you get rid of imaginary numbers).

Captain Chicky
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    Very helpful—thank you! – Leon Aug 18 '23 at 19:35
  • Could it be that the numerator after the substitution is $u^a$ and not $u^{a-1}$? The terms for the Beta function are then $x=2+it$ and $y=-it$. Using the properties of the Gamma function and the second hint I obtain $\varphi_X(t)=-i\pi(1+it)\sinh(\pi t)$. – Leon Aug 19 '23 at 10:35
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    @Leon the numerator has a -1 because of the differential. The substitution $u=e^x$ implies that $dx=du/e^x=du/u$. I just combined this $1/u$ with the top power. So after the substitution, the numerator is indeed $u^a$. Combine this with the differential’s scaling after the substitution and you get $u^{a-1}$ – Captain Chicky Aug 19 '23 at 19:15
  • @Leon I've edited my answer to demonstrate that clearly – Captain Chicky Aug 19 '23 at 19:17
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    Thanks for taking the time to write this out, much appreciated :-) – Leon Aug 19 '23 at 19:56