$\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$?

Question

$\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$? My attempt: $$\int_0^{100}\frac{e^{-x}}{x+100}dx>\int_0^{100}\frac{e^{-x}}{200}dx=\frac{1-e^{-100}}{200}$$ A little bit error. How to amend it?

Answer 1

The idea: The integrand decreases rapidly on the given interval, so the idea is to estimate the integral from below by integrating over a shorter interval $[0, a]$, and then continue with your approach, but with a better bound for the denominator:

For $0 < a < 100$ we can estimate $$I = \int_0^{100}\frac{e^{-x}}{x+100}dx \ge \int_0^a\frac{e^{-x}}{x+100}dx \\ \ge \int_0^a\frac{e^{-x}}{a+100}dx = \frac{1-e^{-a}}{a+100} \, . $$ For $a=4$ this gives $$ I \ge \frac{1-e^{-4}}{104} \approx 0.00943927270299294 $$ which comes fairly close to the result $I \approx 0.009901942286733037$ (obtained by numeric integration with Maxima).

We can also avoid calculating $e^{-a}$ numerically and use $e^a \ge 1+a$ to further estimate $$ I \ge \frac{1-e^{-a}}{a+100} \ge \frac{a}{(a+1)(a+100)} \, . $$ For $a=10$ this gives $$ I \ge \frac{1}{121} \approx 0.008264462809917356 \,. $$

Answer 2

Use $e^u\ge 1+u$ to conclude $$ \frac1{1+u}\ge e^{-u}\implies \frac1{100+x}\ge\frac1{100}e^{-x/100}. $$ Then, again using $1-e^{-u}\ge1-\frac1{1+u}$, $$ \int_0^{100}\frac{e^{-x}}{100+x}dx \ge\frac1{100} \int_0^{100} e^{-\frac{101}{100}x} dx =\frac1{100} \frac{100}{101} \left( 1-e^{-101} \right) \ge\frac1{101}\left(1-\frac1{102}\right)=\frac1{102} $$ which is decidedly larger than $\frac1{200}$.

Answer 3

Using the MatDeck function for numerical integration (Gaussian rule of integration in this case) you can get approximate value of 0.00990194

MatDeck numerical integration - Gaussian rule of integration