How to approximate the integral $\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt$

Question

Suppose we have the following integral \begin{equation} \int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt, \end{equation} where $b$ is a positive constant. It seems very difficult to derive the exact result. So my question is: is there any approximation result? I'm doubting that the approximation result will be just a simple polynomial of the constant $b$. Note that the term $e^{-e^{-t}}$ converge to 1 very fast due to its double exponential structure. Then how to proceed?

UPDATE:I use MATLAB to plot $f_1(b)=\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt$ and $f_2(b)=\log(b)$ in the region $b\in[0, 10]$. The observation is very interesting: they agree very well except a small constant offset...So I'm strongly doubting that this integral is approximately equal to $\log(b)+constant$!

UPDATE 2: please go to Prove $\int_{-b}^{\infty}\log^{\nu}(t+b)e^{-t}e^{-e^{-t}}dt\xrightarrow{b\rightarrow\infty} \log^{\nu}(b)$

Answer 1

We want to study the integral

$$I(b)=\int_{-b}^{\infty}\log\left(b+x\right)e^{-x}e^{-e^{-x}}dx$$

for the two cases $b\rightarrow\infty$ and $b\rightarrow 0^+$.

Let's start with the first one. We write

$$ I(b)=\log(b)\int_{-b}^{\infty}e^{-x}e^{-e^{-x}}+\int_{-b}^{\infty}\log\left(1+\frac{x}{b}\right)e^{-x}e^{-e^{-x}}dx $$

the first integral is elementary. Furthermore we sub $x/b=y$ in the second. This yields

$$ I(b)=\log(b)(1-e^{-e^{b}})+b\underbrace{\int_{-1}^{\infty}\log\left(1+y\right)e^{-b y}e^{-e^{-b y}}dy}_{J(b)} $$

Using Taylor expansion we get

$$ J(b)=\sum_{n\geq1}\frac{(-1)^n}{n}\int_{-1}^{\infty} y^n{e^{- b y}} e^{- e^{- b y}}dy $$

by using the fact that $e^{-e^{-b y}}\le 1$ we may show that the constituents of the above sum are bounded by terms of $\mathcal{O}(b^{-n-1})$ so to the first term will yield the dominant correction to the $\log(b)$ term.

$$ J(b)\sim\int_{-1}^{\infty} y{e^{- b y}} e^{- e^{- b y}}dy+\mathcal{O}(b^{-3}) $$

now substitue $e^{-by}=\xi$ and use the defintion of the exponential integral to show that ($\gamma$ is the Euler-Marschoni constant)

$$ J(b)\sim\frac{\gamma+b e^{-e^{b}}-\text{Ei}{(-e^b)}}{b^2}+\mathcal{O}(b^{-3})\sim \frac{\gamma}{b^2}+\mathcal{O}(b^{-3}) $$

and therefore

$$ I(b)\sim\log(b)+\frac{\gamma}{b}+\mathcal{O}(b^{-2})\quad\text{as}\quad b\rightarrow\infty $$

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As $b\rightarrow 0$ we rewrite (because i'm in a hurry this part will be a bit more sketchy)

$$ I(b)=e^{b}\int_0^{\infty}\log(y)e^{-y}e^{-e^{b}e^{-y}} $$

For small $b$ we might write $e^{b}=1+b+\mathcal{O(b^2)}$

$$ I(b)=(1+b)\int_0^{\infty}\log(y)e^{-y}e^{-e^{-y}}(1-be^{-y})+\mathcal{O}(b^2) $$

or

$$ I(b)=C+b(C-D)+\mathcal{O}(b^2)\quad\text{as}\quad b\rightarrow 0^+ $$

where $C=\int_0^{\infty}\log(y)e^{-y}e^{-e^{-y}}$ and $D=\int_0^{\infty}\log(y)e^{-2y}e^{-e^{-y}}$ are contstants which have to determined numerically $(C\approx -0.155 ,D\approx0.262)$

Answer 2

This is not an answer, I presume.

Computing the integral for $1\leq b\leq 50$ and playing with the numbers, I had a quite good fit $(R^2=0.999999)$ using $$y=0.979379 \log (b+0.131363)+0.085287$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.085287 & 0.003956 & \{0.0773,0.0932\} \\ b & 0.979379 & 0.001140 & \{0.9771,0.9817\} \\ c & 0.131363 & 0.007155 & \{0.1170,0.1458\} \\ \end{array}$$

Edit

There are many values of $b$ which makes the numerical integration quite difficult and probably not very accurate. What bothers me more is that, if i cover the range $0\leq b\leq 50$, the resulting fit is quite different $(R^2=0.999742)$ $$1.06113 \log (x+0.939639)-0.222393$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -0.222393 & 0.049874 & \{-0.3227,-0.1221\} \\ b & +1.061130 & 0.014504 & \{+1.0320,+1.0903\} \\ c & +0.939639 & 0.074340 & \{+0.7901,+1.0892\} \\ \end{array}$$

Update

As I wrote, I faced quite serious difficulties with the direct numerical integration. Interesting, may be, is to notice that all of them disappear if (as you suggested) we compute $$\int_{-b}^\infty \log (b+t)\,e^{-t-e^{-t}} dt=\int_{-b}^0\log (b+t)\,e^{-t-e^{-t}} dt+\int_0^\infty \log (b+t)\,e^{-t-e^{-t}} dt$$ For the range $0\leq b\leq 200$, each integral seems to follow the same $\alpha+\beta \log(b+\gamma)$ with very high values of the $R^2$ (always larger than $0.999$ and the results are $$\int_{-b}^0\log (b+t)\,e^{-t-e^{-t}} dt\approx 0.393913 \log (x+0.806893)-0.126909\qquad R^2=0.999551$$ $$\int_0^\infty \log (b+t)\,e^{-t-e^{-t}} dt\approx 0.622688 \log (x+0.767432)+0.047395\qquad R^2=0.999995$$ We can notice the similarity of the constant terms in the logarithm and, as a good compromise, a global curve fit leads to $$\int_{-b}^\infty \log (b+t)\,e^{-t-e^{-t}} dt\approx 1.01687 \log (x+0.786348)-0.0808138\qquad R^2=0.999959$$

Answer 3

Hint. One may write $$ \begin{align} \int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt&\stackrel{u=t+b}{=}e^b\int_{0}^{\infty}\log(u)e^{-u}e^{-e^be^{-u}}du \\\\&=e^b\int_{0}^{\infty}\log(u)e^{-u}\sum_{n=0}^\infty \frac{(-1)^n}{n!}e^{nb}e^{-nu}du \\\\&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}e^{(n+1)b}\int_{0}^{\infty}\log(u)e^{-(n+1)u}du \\\\&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}e^{(n+1)b}\left(-\frac{\gamma}{n+1}-\frac{\log(n+1)}{n+1} \right) \\\\&=-\gamma \left(e^{-e^b}-1\right)+\sum_{n=1}^\infty \frac{(-1)^n}{n!}\cdot \log n \cdot e^{nb} \tag1 \end{align} $$ where $\gamma$ is the Euler–Mascheroni constant and where the latter infinite sum, being a standard alternating series, satisfies $$ \left|\sum_{n=1}^\infty\frac{(-1)^n}{n!}\cdot \log n \cdot e^{nb}\,-\,\sum_{n=1}^N\,\frac{(-1)^n}{n!}\cdot \log n \cdot e^{nb}\right|\le \left|\frac{\log N\cdot e^{Nb}}{N!} \right| \tag2 $$ and is then easy to approximate.

From $(1)$, one may deduce that a closed form can be given in terms of the generalized Bell polynomials:

$$ \int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt=-\gamma \left(e^{-e^b}-1\right)+e^{-e^b}\left.\frac{\partial}{\partial n}\text{Bell}(n,-e^b)\right|_{n=0}. \tag3 $$