Solving two integrals

Question

Let $f(x) = x^{-2}e^{-x}$ and $g(x) = 2x^{-3}e^{-x}$ . Find $\int f(x)dx$ and $\int g(x)dx$ . I tried to use substitution and integration by parts but didn't help .

Answer 1

Integration by parts should work:

For the first integral, let $u = e^{-x}$ then $du = -e^{-x}dx$. Further let $dv = x^{-2}$ then $v = -1/x$. Thus we have

\begin{align*} -\frac{e^{-x}}{x} - \int \frac{e^{-x}}{x}\\ \Rightarrow -\frac{e^{-x}}{x} + E_1(x) \end{align*}

where $$E_1(x) = \int -\frac{e^{-x}}{x}$$

Now for the case of $E_1(x)$ we can do integration by parts again and let $u = -e^{-x}$ then $du = e^{-x}dx$. Further, let $dv = \frac{1}{x}$ then $v = \ln(x)$. Thus we have for $E_1(x)$

$$-e^{-x}\ln(x) - \int \ln(x)e^{-x}dx$$

This will go on indefinately which is why I stopped at the latter, unless you are given bounds on the integrals there will not be much of an analytical solution.

Answer 2

I will start by working on the $\int x^{-2}e^{-x} dx$ : Use integration by parts with $u = e^{-x} \implies du = -e^{-x} dx$ and $dv = x^{-2} \implies v = -1/x.$

Hence, $\int x^{-2}e^{-x} dx = -\frac{e^{-x}}{x} - \int \frac{e^{-x}}{x} dx$

Now just use another application of integration by parts on the last integral and you are done.

Answer 3

Use integration by parts: $∫uv' = uv - ∫u'v$ and define $u=e^{-x}$ and $v'=x^{-2}$. In a first step we get $ -e^{-x}x^{-1} + ∫ -e^{-x}x^{-1} $. Now note that $E_1(x) = ∫ -e^{-x}x^{-1}$ is a special integral called "exponential integral" (see https://en.wikipedia.org/wiki/Exponential_integral). The solution to your problem is then: $E_1(x) - e^{-x}x^{-1}$

Answer 4

In General: For a non-integer $n>0$, we can evaluate the following integral $$I_n=\int x^{-n}e^{-x}dx$$ $$dv=x^{-n}dx\qquad u=e^{-x}\\v=\frac1{1-n}x^{1-n}\qquad du=-e^{-x}dx$$ $$I_n=\frac1{1-n}e^{-x}x^{1-n}+\int x^{1-n}e^{-x}dx$$ $$I_n=\frac{e^{-x}x^{1-n}}{1-n}+I_{n-1}$$ $$I_n=\frac{e^{-x}x^{1-n}}{1-n}+\frac{e^{-x}x^{2-n}}{2-n}+I_{n-2}$$ Blah blah blah... $$I_n=\sum_{m=0}^{\infty}\frac{e^{-x}x^{m-n}}{m-n}$$ $$I_n=-e^{-x}\sum_{m\geq0}\frac{x^{m-n}}{n-m}$$