What does it mean when an integral has limits at infinity? These integrals are ‘improper!’ In this lesson, learn how to treat infinity as we study the so-called improper integrals.

How Do We Get an Improper Integral?

Have you ever thrown darts at a dartboard? Maybe you’re good and you get really close to the bull’s-eye.

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Or maybe you’re more like me and throw them way off course. It’s almost like you let go of the dart too soon, and it ends up behind you.

Graph for the dartboard problem

Divide the large integral into two smaller ones

tangents rather than just x.

But let’s go back for a minute to trigonometry. If I have cosine squared of theta plus sine squared of theta equals 1, that’s a very standard trig identity that you should all remember. If I take that equation and divide each term by cos^2 of theta, then the first term becomes 1. The second term – which is sin^2 / cos^2 – is the same thing as tan^2, because the tangent of an angle equals the sine of the angle divided by the cosine of the angle. The term on the right side, 1, when I divide it by cos^2 of theta, well that’s 1 over the cosine of theta, all squared, which is the definition of sec^2.

As b goes to infinity, the arctan of b goes to pi/2 because the tangent of pi/2 goes to infinity

So I know that 1 + tan^2 of theta=sec^2 of theta.

That means that this entire bit here simplifies to just sec^2 of theta. So if I plug this in, I get 1 / (sec^2 of theta) * sec^2 of thetadtheta. These sec^2 of thetas cancel out, and I end up with the integral of dtheta.

Well that’s just theta + some constant C. I can now plug in my substitution to make this final integral theta + C, so I can put it in terms of x. To do that, let’s solve x=tan(theta) for theta. So I take the inverse tangent, the arctan, of both sides and I end up with the arctan of x = theta. So my integral, 1 / (1+x^2) * dx = arctan x + C.

Solving an Improper Integral

So let’s go back to our original problem. We’ve divided up our big, nasty minus infinity to infinity integral into two smaller integrals – one from 0 to infinity and one from minus infinity to 0. Both of them have the integrant 1 / (1+x^2). I know that the integral 1 / (1+x^2) * dx = arctan x + C. So let’s see if I can use all of that knowledge.

Let’s focus on this first integral. The integral from 0 to infinity of 1 / (1+x^2) * dx is an improper integral, so I’m going to replace the infinity sign with b and I’m going to take the limit as b goes to infinity.Let’s write this out. The limit as b goes to infinity, from 0 to b, of 1 / (1+x^2) * dx. We know this integral because we just found the indefinite integral, 1 / 1+x^2. So let’s plug in arctan x for this integral from 0 to b.

I get the limit as b goes to infinity of arctan x evaluated from 0 to b. So remember, ‘evaluated from 0 to b‘ means that we’re going to take the arctan of b and subtract from it the arctan of 0. When I do that, I get the limit as b goes to infinity of arctan of b – arctan of 0. As b goes to infinity, the arctan of b goes to pi / 2. Why is this? It’s because the tangent of pi / 2 goes to infinity.

The arctan of 0 is 0, because the tangent of 0 is 0. Remember, arctan is just the opposite, if you will, of the tangent. So my integral from 0 to infinity of 1 / (1+x^2) * dx = pi / 2.So we’ve got the first of my two integrals. Let’s find the second, from minus infinity to 0. We’re going to do the same thing here. First we’re going to replace minus infinity with b.

We’re going to take the limit as b goes to minus infinity of our integral from b to 0 of 1 / (1+x^2) * dx. Our integral is still going to be the arctan of x. So we have the limit as b goes to minus infinity of arctan of x from b to 0; that is, the limit as b goes to minus infinity of the inverse tangent at 0 minus the inverse tangent evaluated at b.

Remember that the inverse tangent is the arctan. Well, the arctan of 0 is equal to 0. As b goes to minus infinity, the arctan of b goes to (-pi) / 2. That’s because the limit, as we go to (-pi) / 2 of the tangent is just minus infinity.

Solution for the improper integral problem

Okay, so I end up with 0 – (-pi) / 2, which just gives me pi / 2.

Plug that in for the second part of my integral, and I get (pi / 2) + (pi / 2). Of course, if you add these, you end up with pi. But let’s stop for a second.

Does it make sense that what’s on the left side, here, is equal to what’s on the right side, here? If you look at the graph, they are the same on either side. So if you put a mirror on the y axis, you would get the same graph no matter what side you were looking at. The left side, where x is negative, and the right side, where x is positive, are mirror images of one another. So it does make sense that both sides would give you the same value for the integral.

Lesson Summary

So this is one example of how to solve improper integrals. Improper integrals are integrals where you’ve got infinity somewhere, either positive or negative infinity. You solve them by splitting up the integrals if you need to, as in our example, and you replace infinity with a variable we call b.

Then you’re going to take the limit as b approaches infinity of your integral.If the limit as b goes to infinity is infinity, then your integral does not converge, so you can’t actually give it a single number. But if you can take the limit as b goes to infinity and get an actual number, then you can solve these improper integrals because they converge.