Can $ \int \frac{1}{x^5} dx $ be used to approximate $ \int \frac{1}{x^5+0.001} dx $?

Question

The following similar indefinite integrals both have very different answers:

$$ \int \frac{1}{x^5} dx $$

$$ \int \frac{1}{x^5+0.001} dx $$

Is there way to approximate an answer to the second integral, using the answer to the first integral?

Answer 1

Yes.

Considering $1/(x^5+t)$ as a function of $t$ gives a Taylor Series of: $$1/(x^5+t)=1/x^5 - t/(x^5)^2 ...$$

Substituting gives $$\int 1/(x^5+t) dx=\int 1/x^5 dx - t \int 1/(x^5)^2 dx ... = \int 1/x^5 dx + t x^{-9}/9 ... $$

Answer 2

For indefinite integrals, no, not in general. Indefinite integration is a tricky business and you have to be much more precise about what type of result you're looking for. It's much easier to be precise with definite integration.

For definite integrals, yes, with some restrictions. For fixed constants $0 <a<b$ we can consider the integral $$ I(\varepsilon) = \int_a^b \frac{1}{x^5 + \varepsilon}dx $$ where $\varepsilon$ is a small positive constant. (The condition $a>0$ is to avoid the singularity of $1/x^5$ at $x=0$, but we could have done $a<b<0$ as well. For this problem we can also take $0<a<b=\infty$, though this isn't always possible for other integrals.) Then we would be interested in the value of $I(0.001)$.

We can get an approximate answer by looking for an asymptotic expansion of $I(\varepsilon)$ as $\varepsilon \to 0$. In general there are lots of techniques that go into computing an asymptotic expansion. In this case we can get the first few terms by computing the Taylor series of $f(x;\varepsilon) = 1/(x^5+\varepsilon)$ with respect to $\varepsilon$ near $\varepsilon = 0$, which looks like $$ f(x;\varepsilon) = \frac{1}{x^5} - \frac{1}{x^{10}}\varepsilon + \frac{1}{x^{15}}\varepsilon^2 + \cdots $$ We can then obtain an asymptotic series expansion of $I(\varepsilon)$ by integrating term by term: $$ I(\varepsilon) \sim \int_a^b \frac{1}{x^5}dx - \varepsilon\int_a^b \frac{1}{x^{10}}dx + \varepsilon^2\int_a^b \frac{1}{x^{15}}dx + \cdots $$ which is valid as $\varepsilon\to 0$. This means that if you take any finite number of terms on the right hand side, then the resulting expression will be a good approximate to $I(\varepsilon)$ for small values of $\varepsilon$. You can check with a computer that this is the case, as I have done here:

Here the horizontal axis is $\varepsilon$, ranging over $\varepsilon\in[0,1]$. We set the limits of integration to $[a,b]=[1,2]$ just for illustration. We plot the values of $I(\varepsilon)$ in red, the two-term expansion $$ \int_1^2\frac{1}{x^5}dx - \varepsilon\int_1^2\frac{1}{x^{10}}dx $$ in blue, and the three-term expansion

$$ \int_1^2\frac{1}{x^5}dx - \varepsilon\int_1^2\frac{1}{x^{10}}dx + \varepsilon^2\int_1^2 \frac{1}{x^{15}}dx $$ in green. We can see that as $\varepsilon\to 0$, the blue and green curves form better and better approximations to the red curve, and the green curve becomes a good approximation faster than the blue curve, as expected.