# What are Improper Integrals? [Explained Easily with Examples]

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Improper Integrals are those integrals where infinity occurs atone or both upper and lower limits of integration while integrating a function f(x).

In simple words, integrals that respond to infinity are called improper integrals. In math, infinity is something that doesn’t have any limits. The area under a curve is the Integral and integrating it gives us the area under the curve of the function f(x) by adding various small parts of that area. The symbol looks like S and on its top, and bottom limits are defined, which gives an idea of the upper limit and lower limit of the function f(x).

## Improper Integrals Explained

To get a basic idea of how Improper Integrals looks like we can see that basically, an improper integral is a limit similar to the form:

Now, this function seems small and easy, but It just cannot be solved by integrating as normal and put the infinity at the place to get an answer as infinity is not a real number. To solve this function, we need to think this function as an area problem and find the area of this function on the interval (1, ∞). This the representation of Improper Integrals. But to solve this function, we will take interval as (1, c) where c >1 and ‘c’ is infinite.

So, this is how you can solve this kind of problems in general terms. The infinity is replaced by a variable (in this case is c), then we do an integral part, and then the limit of the result is taken as ‘c’ tends to infinity. You can also notice that the area under the curve of this function is not infinity, but this will not always be the case, and not all areas at the infinite intervals will give infinity.

## Two Major Parts of Improper Integrals

Further, we will take it into two parts:

1. Convergent – When a limit is a finite number, then it is called convergent.
2. Divergent – When the limit is infinity or doesn’t exist, then it is divergent.

The function we solved earlier is convergent as it has a finite limit. Now, let’s take an example of a divergent function.

Now, this function is something more complicated than the earlier one. We can see that both the limits of the function tend to infinity whereas in earlier function only one limit tends to be infinite. These limits exaggerate the function and make it tougher than the single infinity function. So, to solve this, you have to split the integral into two parts turning both the parts into a limit. Splitting the integral at x=0 is easy because dealing with zero in integration is quite simple than dealing with those complex numbers. Zero is also a preferred choice because it simply lies between (-∞) to (∞) but this theory is not exactly correct because there is no middle point between -infinity and infinity so we are only assuming that zero lies at the middle in between -infinity and infinity.

### Addition After Solving Individual Parts

Taking a note, if either part of the integral diverges then the whole part diverges. We can’t get the infinity for one integral part and a negative infinity for other integral parts and then summing them up to get the zero.
These three processes will ease an integral part of multiple infinite limits.

## Explained with Examples

Let’s take another example of Improper Integrals.

In this function, we have a lower limit as -infinity and upper limit as 0. We also have root in the denominator part, so we will solve this function without breaking the roots.

Let’s solve the equation:

So, we get infinity as the limits for this function, so it is divergent.

Let us take one more example.

Now, this function has sine terminology so we will proceed with this function as it is.

As this limit doesn’t exist so, this is also divergent.

In most of the examples of the infinite intervals the limit either exists or is infinite. But there are some limits too that don’t exist as the last example.

There is also one more type of improper integral, which is Discontinuous Integrand.

In this case, the integrand can be discontinuous within the limits of the integration. This tells us that the integrand has the vertical asymptote or an infinite discontinuity which is at the limits of the integration or maybe at some other point in the interval of the integration.

## Properties of Integrals having Discontinuities

The integrals having discontinuities holds some properties which are as follows.

• If f is discontinuous at x=a in the interval [a, b] then the improper integral is as:

• Here if f is discontinuous at x=b over the interval [ a, b] then the improper integral is as:

• If the integrand f is discontinuous at x=c over the interval [ a, b] then the integral is as:

Let’s take an example:

For this discontinuous integrand function, we will solve it just like earlier functions.

So, the integral converges to

This is basically how you solve Improper Integrals.

Hello folks. I have started this website for all of you who have troubles grasping those tough mathematical concepts. I personally have faced a lot of issues and that is the sole motivation for starting this. Hope it helps you guys. Cheers!