Suppose that we have the following integral \begin{equation} \int_{-b}^{\infty}(t+b)^{\nu}e^{-t}e^{-e^{-t}}dt, \end{equation} where $b$ is a positive constant and $\nu$ is any number (real or complex). How can we get an approximated result? I'm doubting that the form $b^{c_1\nu+c_2\nu^2}$ can approximate it very well.

  • $\begingroup$ I suppose the integral goes as $b^{\nu}$ if $b$ gets big. Furthemore i'm almost certain the corrections will be of order $ b^{-1}$ $\endgroup$
    – tired
    Nov 14, 2016 at 15:48
  • $\begingroup$ Thank you very much for your comment. I think the asymptotic $b^{\nu}$ only applies when $\nu$ is not very large. You can try to sweep $\nu\in[-j20,j20]$, where $j=\sqrt{-1}$. $\endgroup$ Nov 14, 2016 at 17:13
  • $\begingroup$ you mean another question i asked? math.stackexchange.com/questions/2010717/… $\endgroup$ Nov 14, 2016 at 17:33
  • $\begingroup$ yes, I just saw it! it is correct! $\endgroup$ Nov 14, 2016 at 17:36

1 Answer 1


Not easy.

Since you want an approximation, we can do as follows.

The first thing to analyze is the trivial case in which $b = 0$. This will help us later. In this case you have

$$\int_0^{+\infty} t^{\nu} e^{-t} e^{-e^{-t}} \ \text{d}t$$

Since the range of integration is $\mathbb{R}^+$ you are allowed to use Taylor Series for $e^{-e^{-t}}$, and we get:

$$\sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!} \int_0^{+\infty} t^{\nu} e^{-t} e^{-kt}\ \text{d}t = \sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!} \int_0^{+\infty} t^{\nu} e^{-t(1+k)}\ \text{d}t$$

This is a standard integration which gives you

$$\int_0^{+\infty} t^{\nu} e^{-t(1+k)}\ \text{d}t = (1+k)^{-1-\nu}\ \Gamma(1+\nu)$$

Where $\Gamma(a)$ represent the well known Gamma Function.


$$ \sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!} (1+k)^{-1-\nu}\ \Gamma(1+\nu) = \Gamma(1-\nu) \sum_{k = 0}^{+\infty}\frac{(-1)^k}{(k+1)!} (1+k)^{-\nu}$$

Unfortunately, there is no sufficient data to be able to tell if the series does converge and where, because we know nothing about $\nu$. If it's real, complex, imaginary, greater than $2$, between $0$ and $1$ and so on.

I can plot you this series, for $k$ from $0$ to $100$, and $\nu$ from $-8$ to $+8$, to show you the partial behavior, if it helps:

enter image description here

Now we come to a more general case, if it is possible.

When $b\neq 0$ we have your integral.

$$\int_{-b}^{+\infty} (t+b)^{\nu} e^{-t} e^{-e^{-t}}\ \text{d}t$$

General Integral

For this integral, which is a beast, it's conveniente to split it into two parts.

The first one:

$$\int_0^{+\infty} (t+b)^{\nu} e^{-t} e^{-e^{-t}}\ \text{d}t$$

"can be evaluated" by a giant use of Taylor series for both $e^{-t}$ and $e^{-e^{-t}}$. And it's not over. So, using the series we get (easy to verify):

$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \frac{(-1)^k (-1)^j}{k!\ j!} \int_0^{+\infty} (t+b)^{\nu} t^k\ e^{-tj}\ \text{d}t$$

Again we use Taylor series for the last term $e^{-tj}$, getting

$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \sum_{n = 0}^{+\infty} \frac{(-1)^k (-1)^j (-1)^n}{k!\ j!\ n!} j^n\int_0^{+\infty} (t+b)^{\nu} t^k\ t^n\ \text{d}t$$

The integral now is

$$\int_0^{+\infty} (t+b)^{\nu} t^{k+n}\ \text{d}t = b^{1 + \nu + n + k} \frac{\Gamma(-1-\nu - j - n)\Gamma(1 + j + n)}{\Gamma(-\nu)}$$

And so you start seeing how difficult is to get an approximate form, since this result only holds for $\Re(\nu + j + n) >-1$. But it's something, right?

So this part will give you in the end

$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \sum_{n = 0}^{+\infty} \frac{(-1)^k (-1)^j (-1)^n}{k!\ j!\ n!} j^n b^{1 + \nu + n + k} \frac{\Gamma(-1-\nu - j - n)\Gamma(1 + j + n)}{\Gamma(-\nu)}$$

Good luck in summing that. Even if you may try with the first $3-4$ terms each series, that would be simple, actually. I cannot say anything about the convergence of those series, though. The parameter $\nu$ is too general.

For what concerns the second integral, we have

$$\int_{-b}^0 (t+b)^{\nu} e^{-t} e^{-e^{-t}}\ \text{d}t$$

What we can do here are two passages: the first is to use the Binomial theorem for $(t+b)^{\nu}$ term, and the second is again a Taylor series for the second exponential, allowed because


has two parts: the "higher" $e^{-t}$, which becomes great for $t < 0$, and the second part $e^{-e^{-t}}$ which becomes very small for $e^{-t}$ large.


$$(t+b)^{\nu} = \sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} t^m$$

$$e^{-e^{-t}} = \sum_{p = 0}^{+\infty} \frac{(-1)^p}{p!} e^{-pt}$$

Putting all together you get

$$ \sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} \sum_{p = 0}^{+\infty} \frac{(-1)^p}{p!} \int_{-b}^0 t^m e^{-t} e^{-tp}\ \text{d}t$$

The very last integral is trivial

$$ \int_{-b}^0 t^m e^{-t} e^{-tp}\ \text{d}t = \frac{(-1)^m }{(-p-1)^{m+1}} (\Gamma (m+1)-\Gamma (m+1,-b (p+1)))$$

Where the last function is the Incomplete Gamma Function. You can manipulate a bit the minus signs, and in the end with the previous sums you have:

$$ \sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} \sum_{p = 0}^{+\infty} \frac{(-1)^{p+1}}{(p+1)!} (\Gamma (m+1)-\Gamma (m+1,-b (p+1)))$$

The final Union

Summing this part to the previous and we get the (actually not so approximate, but rather quite exact for the conditions we established about the parameters) solution:

$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \sum_{n = 0}^{+\infty} \frac{(-1)^k (-1)^j (-1)^n}{k!\ j!\ n!} j^n b^{1 + \nu + n + k} \frac{\Gamma(-1-\nu - j - n)\Gamma(1 + j + n)}{\Gamma(-\nu)} + \\\\ +\sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} \sum_{p = 0}^{+\infty} \frac{(-1)^{p+1}}{(p+1)!} (\Gamma (m+1)-\Gamma (m+1,-b (p+1)))$$

  • $\begingroup$ I did not go through your steps but I admire your patience with this monstreous problem ! $\endgroup$ Nov 14, 2016 at 11:40
  • $\begingroup$ @ClaudeLeibovici Thanks again Mr! I just had some fun XD $\endgroup$
    – Laplacian
    Nov 14, 2016 at 12:14
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    $\begingroup$ @AlanTuring, It's really interesting to discuss this problem with you. In my last comment, I mean that the Taylor expansion $e^{-e^{-t}}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}e^{-nt}$ does not work very well especially when $t$ is a very small negative number, for example, $t=-5$. Actually, approximating a very small number, e.g., $e^{-e^{5}}=3.5\times10^{-65}$ requires a very large number of terms, this may be even not very practical. $\endgroup$ Nov 14, 2016 at 13:59
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    $\begingroup$ You can find this from some double exponential distributions, for example referring to this en.wikipedia.org/wiki/Gumbel_distribution $\endgroup$ Nov 14, 2016 at 14:16
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    $\begingroup$ kawofengche it looks like you did virtually nothing, and then, after Alan Turing provided you with a very detailed and thorough answer (No doubt he spent far more time on answering your question, than you put into it. And then, you nit-pick that answer? Thee who hath not worked ought not judge others who doeth thy work. $\endgroup$
    – amWhy
    Nov 15, 2016 at 0:25

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