# The Unapologetic Mathematician

## Improper Integrals I

We’ve dealt with Riemann integrals and their extensions to Riemann-Stieltjes integrals. But these are both defined to integrate a function over a finite interval. What if we want to integrate over an infinite ray, like all positive numbers?

As a specific example, let’s consider the function $f(x)=\frac{1}{x^2}$, and let it be defined on the ray $\left[1,\infty\right)$. For any real number $b>1$ we can pick some $a'>a$. In the interval $\left[1,b'\right]$ the function $f$ is continuous and of bounded variation (in fact it’s decreasing), and so it’s integrable with respect to $x$. Then it’s integrable over the subinterval $\left[1,b\right]$. Why not just start by saying it’s integrable over $\left[1,b\right]$? Because now we have a function on $\left[1,b'\right]$ defined by

$\displaystyle F(x)=\int\limits_1^x\frac{1}{t^2}dt$

Since $t$ is differentiable and $\frac{1}{t^2}$ is continuous at $a$, we see that $F$ is differentiable here, and its derivative is $F'(b)\frac{1}{b^2}$. This result is independent of the $b'$ we picked.

Since we can do this for any $b>1$ we get a function $F(b)$ defined for $b\in\left(1,\infty\right)$. Its derivative must be $\frac{1}{b^2}$, and we can check that $\frac{-1}{b}$ also has this derivative, so these two functions can only differ by a constant. Clearly we want $F(1)=0$, since at that point we’re “integrating” over a degenerate interval consisting of a single point. This fixes our function as $F(b)=1-\frac{1}{b}$.

Now our question is, what happens as we take $a$ to get larger and larger? Our intervals $\left[1,b\right]$ get bigger and bigger, trying to fill out the whole ray $\left[1,\infty\right)$. And for each one we have a value for the integral: $1-\frac{1}{b}$. So we take the limit as $b$ approaches infinity: $\lim\limits_{b\rightarrow\infty}1-\frac{1}{b}=1$. This will be the value of the integral over the entire ray.

We turn this rubric into a definition: given a function $f$ that is integrable with respect to $\alpha$ over the interval $\left[a,b\right]$ for all $b>a$, we can define a function $F$ on $\left[a,\infty\right)$ by

$\displaystyle F(b)=\int\limits_a^bfd\alpha$

We define the improper integral to be the limit

$\displaystyle\int\limits_a^\infty fd\alpha=\lim\limits_{b\rightarrow\infty}\int_a^bfd\alpha$

if this limit exists. Otherwise we say that the integral diverges.

We can similarly define improper integrals for leftward rays as

$\displaystyle\int\limits_{-\infty}^bfd\alpha=\lim\limits_{a\rightarrow-\infty}\int_a^bfd\alpha$

And over the entire real line by choosing an arbitrary point $c$ and defining

$\displaystyle\int\limits_{-\infty}^\infty fd\alpha=\int\limits_{-\infty}^cfd\alpha+\int\limits_c^\infty fd\alpha$

That is, we take the two bounds of integration to go to their respective infinities separately. It must be noted that the limit where they go to infinity together:

$\displaystyle\int\limits_{-\infty}^\infty fd\alpha=\lim\limits_{b\rightarrow\infty}\int\limits_{-b}^bfd\alpha$

may exist even if the improper integral diverges. In this case we call it the “Cauchy principal value” of the integral, but it is not the only justifiable value we could assign to the integral. For example, it’s easy to check that

$\displaystyle\lim\limits_{b\rightarrow\infty}\int\limits_{-b}^bxdx=\lim\limits_{b\rightarrow\infty}0=0$

so the Cauchy principal value is ${0}$. However, we might also consider

$\displaystyle\lim\limits_{b\rightarrow\infty}\int\limits_{1-b}^{1+b}xdx=\lim\limits_{b\rightarrow\infty}2b$

which diverges.

April 18, 2008 - Posted by | Analysis, Calculus

1. [...] Tests for Improper Integrals We have a few tests that will come in handy for determining if an improper integral converges. In all of these we’ll have an integrator on the ray , and a function which is [...]

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2. [...] on the ray , and be any function integrable with respect to through the whole ray. Then if the improper integral converges, then so does [...]

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3. Is there some significance attached with the Cauchy Principal Value? I mean since it has been given a name, it most have encountered at more places.

Comment by Nilay | April 23, 2008 | Reply

4. I’m not sure what I’ll do with it yet, but I may as well give its name now, and if it comes up later I can refer back to it.

Comment by John Armstrong | April 23, 2008 | Reply

5. [...] the particular case of an improper integral, we have . Then . Our condition then [...]

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6. [...] Then as we let go to infinity, goes to infinity with it. Thus the sum of the series is the same as the improper integral. [...]

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8. [...] I get the title of the post wrong? No. It turns out that I covered half of this topic two years ago as I prepared to push into infinite series. Back then, I dealt with what happened when we wanted to [...]

Pingback by Improper Integrals II « The Unapologetic Mathematician | January 15, 2010 | Reply