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I want to estimate $$\int_{1/m}^{\infty} y^{-1} e^{-y} dy,$$where $m$ is a positive number.

My Attempt

$$\begin{aligned} \int_{1/m}^{\infty} y^{-1} e^{-y} dy &= -y^{-1}e^{-y}|_{1/m}^{\infty} - \int_{1/m}^{\infty} y^{-2} e^{-y} dy\\ &\leq me^{-1/m}\leq m. \end{aligned}$$

It seems that this estimate is not sharp enough. In fact, my intuition told me that the result may involve $\ln m$, but I cannot figure it out. Could you give me a hint? Any insight or help is appreciated.

Louise Wing
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  • Welcome to MSE. Here's how to ask a good question. Follow these guidelines to get help in this forum. It's particularly important that you share your own work and thoughts on the problem to show that you have made a serious effort by yourself before asking for help, and you're not just trying to get others to solve it for you. – jjagmath Oct 25 '23 at 11:40
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    Hint: the integral from $1$ to infinity is just a constant. For the integral from $1/m$ to $1$, you can substitute $y=1/t$ and expand the exponential as a power series. – Luca Armstrong Oct 25 '23 at 12:26

2 Answers2

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Note that Ei(x)=\int_{-x}^{\infty}\frac{e^{-x}}{x}dx

the x value in this function is \frac{-1}{m}

Thus to find any value of this integral you can use the function

I=Ei(\frac{-1}{m}) , where Ei(x) is the exponential integral

Then you can use this infinite sum

Ei(x)=-\gamma-\ln(x)-\sum_{1}^{\infty}(\frac{(-x)^{k}}{kk!}) , where \gamma is the euler-mascheroni constant

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Using the exponential integral function $$\int y^{-1} e^{-y}\, dy=\text{Ei}(-y)$$ Then, using the incomplete gamma function equivalent $$I=\int_{\frac 1m}^{\infty} y^{-1} e^{-y}\, dy=\Gamma \left(0,\frac{1}{m}\right)$$ and, if $m$ is large $$I=\log (m)-\gamma +\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)\,(n+1)!\,m^{n+1}}$$

Not only it is alternating but moreover, it will converge very fast since $$\left| \frac{a_{n+1}}{a_n}\right|=\frac{n+1}{m (n+2)^2}\sim \frac 1{mn}$$

This gives very simple bounds (as tight as you wish). The simplest would be $$\log (m)-\gamma +\frac{1}{m}-\frac{1}{4m^2} <I < \log (m)-\gamma +\frac{1}{m}$$

Moreover, for calculation purposes, you can know in advance the number of terms to be added for a given accuracy. Writing the infinite sammation as $$\sum_{n=0}^p\frac{(-1)^n}{(n+1)\,(n+1)!\,m^{n+1}} +\sum_{n=p+1}^\infty \frac{(-1)^n}{(n+1)\,(n+1)!\,m^{n+1}}$$ you want to know $p$ such that $$R_p=\frac 1 {(p+2) (p+2)! \,m^{p+2} }\leq 10^{-k}$$ or, to make it simpler, $$ (p+3)! \,m^{p+2} \geq 10^k$$

If you look at this old post of mine, using @robjohn's approximation, we have, as a real, $$p \sim \frac 1 m \,e^{1+W(t)} -\frac 7 2 \qquad \text{where} \qquad t=\frac m {2e}\log \left(\frac{10^{2 k}}{2 \pi m}\right)$$

Claude Leibovici
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