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My integration techniques are rusty. If I wish to integrate the following function, I can do it by using integration by parts twice, but is this really the simplest way?

I wish to integrate the function $f(x) = (x^2 - 2x + 1)(e^{-x})$. Integrating by parts we get $$\int f(x)dx = (x^2 - 2x + 1)(-e^{-x}) - \int(2x-2)(-e^{-x})dx$$ In order to solve the last integral we must again do integration by parts to get $$\int(2x-2)(-e^{-x})dx = (2x-2)(e^{-x})-\int 2(e^{-x})dx$$ The final integral can then be solved directly and we get the final result, which is $$\int f(x)dx = -e^{-x}(x^2+1)$$ But is this the simplest way? Does this mean I need to use integration by parts $n$ times if I have a polynomium of $n$th degree? Or are there some tricks when a polynomium is multiplied by $e^{-x}$?

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    $\begingroup$ The only "trick" I can think of is using a reduction formula but that's derived from integration by parts. $\endgroup$ – Justin Stevenson Dec 10 '18 at 1:28
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A trick for such integrals:

  • If $f(x)$ is of the form $f(x)$ = $e^x$ $(g(x) + g\prime(x))$, then $\int f(x)$ = $e^x g(x)$ + some constant.
  • If $f(x)$ is of the form $f(x)$ = $e^{-x}$ $(g\prime(x) - g(x))$, then $\int f(x)$ = $e^{-x}g(x)$ + some constant.

The given question is of the second form. I am sure you can prove the trick so I am skipping it.

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Integration by parts done fast This method speeds up the process, but other then this I don't know of any 'tricks'

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"Best method" is subjective. But another way is to use Feynman's trick of differentiating under the integral sign.

First, note that the integral is $\displaystyle \frac 1e \int (x-1)^2e^{-(x-1)}dx$

Now consider the integral $g(x,k) = \displaystyle \frac 1e \int e^{-k(x-1)}dx$ and note that $\displaystyle \frac{\partial{^2}}{\partial k^2}g(x,k) = \frac 1e \int (x-1)^2e^{-(x-1)}dx$ when $k=1$.

$g(x,k)$ is trivial to compute, and the twice partial differentiation can be done with product rule without too much fuss.

EDIT (addendum): even if you were to use integration by parts twice, your life would be made easier by doing a substitution $\displaystyle u = x-1$, to make the integral $\displaystyle \frac 1e \int u^2e^{-u} du$, which is more tractable.

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