# Why is it obvious that this integral is undefined?

If we have the following integral:

$$\int_0^1\frac{1}{\left(x+a\right)\left(x+2\right)}\,dx$$

Why is it that it is obvious that this integral will be undefined for $0>a>-1$?

I have a rough idea that one knows that this creates an asymptote at $x=-a$ and this means that if $-a$ lies in the range $0<-a<1$ then you are trying to get the area under something rushing off to an asymptote. Why is it that this is so immediately obvious? Could somebody explain the reasoning to reach the conclusion that it is undefined for the values above to confirm that mine above is correct?

It isn't obvious until you have looked at many of these. I will try to give an informal explanation of why some people would think of it as obvious. It isn't just about the asymptote, but about the order of the asymptote: $x^{-1}$, $x^{-2}$, and $x^{-1/2}$ behave differently around their asymptotes at $x = 0$.

Informally, if we have a "order $1$" asymptote, like $1/x$ has at $x = 0$, then the area trapped by the curve on each side of the asymptote will be infinite. In particular, we have rigorous results such as $\int_0^1 x^{-1}\,dx = \infty$ and $\int_{-1}^0 x^{-1}\,dx = -\infty$.

This leads to an informal rule of thumb which can be applied in this setting to see that the overall integral won't converge. The $1/(x+2)$ part won't affect the behavior between $0$ and $1$ because $1/(x+2)$ is bounded on that interval. The $1/(x+a)$ piece is raised to the power $1$, so it will behave like $1/x$ would at $x = 0$.

The second part of the informal rule of thumb says that if we have a function such as $1/\sqrt{x}$ (or more generally $1/x^c$ for $0 < c < 1$), then the area trapped by the vertical asymptote will be finite. In particular, we rigorously obtain $\int_0^1 x^{-1/2}\,dx = 2$. Similarly, the informal rule of thumb says that $$\int_0^1 \frac{1}{(x+a)^{2/3}(x+2)}\,dx$$ will converge for any value of $a \in [0,1]$. Because of the $2/3$ power, this integral will act like the integral of $x^{2/3}$ does around $x = 0$.

The final part of the informal rule of thumb says that if we have $1/x^c$ for $c \geq 1$, then the area trapped by the vertical asymptote will be infinite. So, rigorously, the integral $\int_0^1 x^{-3}\,dx$ will not exist.

There is one other issue which happens when the asymptote is not at the edge of the interval of integration. In this case, in order for the overall integral to exist, the integrals on both sides of the asymptote must exist, and in particular both of these must be finite. This is why we do not compute $\int_{-1}^1 x^{-1}\,dx$ to be zero.

That is the rule of thumb when there is a vertical asymptote. There is a different rule for horizontal asymptotes. So, for example $\int_2^\infty 1/x^c\,dx$ converges if and only if $c > 1$, while $\int_0^2 1/x^c\,dx$ converges for $0 < c < 1$. (We can convert between the two forms, actually, by thinking about expressing the area as a double integral and changing the order of integration.)

• This may be getting beyond the intent of the question, but in more advanced settings, this kind of phenomenon explains why there are conditions on which $L^p$ spaces are contained in which other $L^p$ spaces. See en.wikipedia.org/wiki/Lp_space#Embeddings for the statement of the result. – Carl Mummert Dec 2 '17 at 21:59

Assuming $a\ne2$,\begin{align}\int_0^1\frac1{(x+a)(x+2)}\,dx&=\frac1{2-a}\int_0^1\left(\frac1{x+a}-\frac1{x+2}\right)\,dx\\ &=\frac1{2-a}\left[\ln(x+a)-\ln(x+2)\right]_0^1\\ &=\frac1{2-a}\left(\ln\left(\frac{1+a}{a}\right)-\ln\left(\frac{3}{2}\right)\right)\end{align} So for $a\in(-1,0)$, you have $\frac{1+a}{a}<0$, and so the integral is undefined in this range.

• $\frac{1}{x+a}$ has a non-integrable singularity at $-a$. Plug $a = -\frac{1}{2}$ into your formula, and it contains a $\ln (-1)$. – Daniel Fischer Dec 2 '17 at 21:42