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Stieltjes integral I

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In this post we would like to illustrate the calculation of an integral using perturbation techniques. We would like to calculate the following Stieltjes integral:

$\displaystyle I = \int_{0}^{\infty} \frac{e^{-t}}{1+t}dt$

This integral can be evaluated exactly (I = 0.5963), but we’ll use perturbation techniques to compare the results obtained using these approximation techniques with the exactly calculated value. In order to do so, we will insert the parameter $\epsilon$ as follows:

$\displaystyle I(\epsilon) = \int_{0}^{\infty} \frac{e^{-t}}{1+\epsilon^2t}dt$

For $\epsilon = 0$ we have $I(\epsilon)$ = 1. This is the solution to the unperturbed problem. We observe that:

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + ...$

$\displaystyle = 1 + x + x^2 + ... + x^n + x^{n+1}(1 + x + x^2 + ...)$

$\displaystyle = 1 + x + x^2 + ... + x^n + \frac{x^{n+1}}{1-x}$

Defining $x := -\epsilon^2 t$ and using $\int_{0}^{\infty} e^{-t} t^n dt = n!$ in the formula above we obtain:

$\displaystyle \frac{1}{1 +\epsilon^2t} = 1 -\epsilon^2 t + \epsilon^4 t^2 + ... + (-1)^n \epsilon^{2n} t^{n} + \frac{(-1)^{n+1} \epsilon^{2n+2} t^{n+1}}{1+\epsilon^2 t}$

$\displaystyle I(\epsilon) = \int_{0}^{\infty} e^{-t}(1 -\epsilon^2 t + \epsilon^4 t^2 + ... + (-1)^n \epsilon^{2n} t^{n} + \frac{(-1)^{n+1} \epsilon^{2n+2} t^{n+1}}{1 + \epsilon^2 t}) dt$

$\displaystyle = \int_{0}^{\infty} (e^{-t} -\epsilon^2 e^{-t} t + \epsilon^4 e^{-t} t^2 + ... + (-1)^n \epsilon^{2n} e^{-t} t^{n} + \frac{(-1)^{n+1} \epsilon^{2n+2} e^{-t} t^{n+1}}{1 + \epsilon^2 t}) dt$