A Family of Integrals

Given that

$\begin{aligned} I_n & = \int_0^1 x^{2019}\ln^2 x\ dx & \small \blue{\text{Let }e^u = x \implies e^u \ du = dx} \\ & = \int_{-\infty}^0 e^{2020 u} u^n \ du & \small \blue{\text{Let }t = -2020 u \implies dt = - 2020\ du} \\ & = \int_0^\infty \frac {(-t)^n e^{-t}}{2010^{n+1}} dt & \small \blue{\text{Gamma function }\Gamma (s) = \int_0^\infty t^{s-1}e^{-t} dt} \\ & = \frac {(-1)^n}{2020^{n+1}} \Gamma (n+1) & \small \blue{\text{Since }\Gamma (n) = (n-1)!} \\ & = \frac {(-1)^nn!}{2020^{n+1}} \end{aligned}$

Therefore, $\displaystyle \sum_{n=0}^\infty \frac 1{I_n} = \sum_{n=0}^\infty \frac {2020(-2020)^n}{n!} = 2020 e^{-2020} = \frac {2020}{e^{2020}} \implies a + b = 2020+2020 = \boxed{4040}$ .

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Reference: Gamma function

Consider the function $I(t)$ given by $I(t) = \int_0^1 x^t\,dx, \,\, t\geq 0$ then via the Leibniz rule of differentiation under the integral sign we may write $\frac{d^n}{dt^n} I(t) = \frac{d^n}{dt^n}\int_0^{1}x^t\,dx = \int_0^1 \frac{\partial^n}{\partial t^n} x^t\,dx = \int_0^1 x^t \ln(x)^{n}\,dx$ Therefore the value we seek is the $n$ th derivative of $I(t)$ evaluated at $t=2019$ . Doing this is simple since $I(t)$ is a simple integration: $I(t) = \int_0^1 x^t\,dx = \frac{1}{t+1}, \,\, t\geq 0.$ Simple patterning will reveal that the $n$ th derivative of $I(t)$ is $\frac{d^n}{dt^n} I(t) = \frac{(-1)^n n!}{(t+1)^{n+1} }$ Therefore $I_n$ is $I^{(n)}(2019)$ , i.e each element in the set $\left\{I_n\right\}$ may be written as $I_n = \frac{(-1)^n n!}{(2020)^{n+1}} \Longrightarrow \frac{1}{I_n} = \frac{(-1)^n 2020^{n+1}}{n!}$ The requested sum is therefore $\sum_{n=0}^{\infty} \frac{1}{I_n} = \sum_{n=0}^{\infty} \frac{(-1)^n 2020^{n+1}}{n!} = 2020\sum_{n=0}^{\infty} \frac{(-1)^n 2020^{n}}{n!}$ Direct comparison with the Taylor series for $e^x$ about $x=0$ : $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ reveals that the requested sum is: $\sum_{n=0}^{\infty}\frac{1}{I_n} = 2020e^{-2020} = \frac{2020}{e^{2020}}.$ Therefore $\boxed{ a+ b = 4040}$