Integral Test

Sequence and series are a building block of any analytical process. A function’s continuity can be easily proved with the help of sequences. In this respect, the integral test is used to find out whether a given series under analysis is in convergence or not. The convergence of series is important when the integral function has the sum of a series of a function. Hence, it is quite essential to check whether a series is in convergence or not in case of some specific functions.

If \[\int_{1}^{\infty} f(x) dx\] coverges, then \[\sum_{n=1}^{\infty} a^{n}\] converges.

If \[\int_{1}^{\infty} f(x) dx\] diverges, then \[\sum_{n=1}^{\infty} a^{n}\] diverges.

Integral Test for Convergence

In the mathematical domain, Integral test for convergence is a technique which is often applied for the purpose of testing an infinite series of non-negative terms for convergence. The method is also known as the Maclaurin-Cauchy test as Colin Maclaurin, and Augustin-Louis Cauchy developed it.

For example, if n is a neutral non-negative number, and it is a monotonically decreasing function, then the function is defined as

f: [n, ∞ ]→ R

In this case, the series will be convergent only if the integral is finite.

Integral Test Conditions

An integral comparison test is carried out mainly for integral terms. For instance, if there are two functions including f(x) and g(x) and g(x) ≥ f(x) on the given interval [c, ∞], then the following conditions should be true:

• In case the term converges, then the term so does.

• In case the term divergences, then the term so does.

Let’s understand the integral test series with the help of an example:

Example: Identify if the given series is convergent or divergent:

In this case, the function that will be used will be

F(x) = 1/ x In x

The function is positive, and if x is made larger, the denominator will be larger, and so the function will be decreasing. Thus, to determine the convergence of the following integral what needs to be done is

\[\int_{2}^{\infty} \frac{1}{x} ln x dx = \lim_{t \rightarrow \infty} \int_{2}^{t} \frac{1}{x} ln x dxu = lnx\]

\[= \lim_{t\rightarrow \infty} (ln(lnx))|_{2}^{t}\]

\[= \lim_{t\rightarrow \infty} (ln(lnt) - ln(ln2))\]

\[= \infty\]

This proves that the integral is divergent and so the series is divergent by the integral test.

Proof of the Integral Test

The integral test proof depends on the comparison test of the series. By now we know that is nothing but a sum of series \[\sum_{m}^{\infty} = N\int_{m+1}^{m} f(t)dt\].

Since “f” here is a monotonically decreasing function, then

f(t) ≤ f(m) for every “t” in [m, ∞]

For m > N, \[\int_{m+1}^{m} f(t)dt \leq \int_{m+1}^{m}f(m)dt = f(m)\]

It means that \[\int_{m+1}^{m} f(t)dt \leq f(m)\]

Since both the quantities are non-negative, a comparison test will be used.

If\[\sum_{m}^{\infty} = Nf(m)\] converges, then \[\sum_{m}^{\infty} = N\int_{m+1}^{m}f(t)dt = \int_{N}^{\infty} f(t)dt\] will also be convergent.

Thus, it is finite.

This, however, is just the first step of the proof.

Now, we again assume that the function f is a monotonically decreasing function, the equation that we get now is,

(m) ≤ f(t) for every “x” in [M, m]

So, \[f(m) = \int_{m-1}^{m} f(t)dt \leq \int_{m-1}^{m} f(t)dt\]

The comparison theorem proves that 

If \[\sum_{m}^{\infty} = N\int_{m}^{m-1} f(t)dt = f(N) + \int_{N}^{\infty} f(t)dt\] converges, then we can say that \[\sum_{m}^{\infty} = Nf(m)\] will also converge. Hence, this proves the second part of the theorem.

The Comparison Test

If f(x) ≥ g(x) ≥ 0, on the interval [a,∞) then,

1. If \[\int_{a}^{\infty} f(x)dx\], converges then so does \[\int_{a}^{\infty}g(x)dx\].

2. \[\int_{a}^{\infty} g(x)dx\], diverges then so does \[\int_{a}^{\infty} f(x)dx\].

A comparison test makes a lot of sense when thinking in terms of area. If the function f(x) is larger than the function g(x) the area under the function f(x) will be larger than the area under the function g(x).

Thus, if the area under the larger function \[\int_{a}^{\infty} f(x)dx\] is finite and converges, then the area under the smaller function, i.e., \[\int_{a}^{\infty} g(x)dx\] will also converge and be finite.

Let’s take an example to understand this better.

Example: Find out whether

I = Converges or diverges

Solution: The function I does not have an expression in terms of elementary function. However

1 ≤ x ⇒ x ≤ x²

The last inequality follows as it is an increasing function

\[0 \leq = |_{1}^{\infty} = 1/e

Since 0 ≤ 1/e, the integral here converges.