Integral Test: Convergence, Divergence, Proof & Conditions

Integral test is used to find whether the given series under analysis is convergence or divergence. The convergence of a series is more significant when the integral function has the sum of the series of functions.

• Therefore, while dealing with some specific functions in sequences and series, it is necessary to check whether the given series is convergent. There are few conditions and tests to prove the convergence of a p-series.
• For the analysis process, sequences and series act as a building block and by using sequences, it can be easily proven if the function is continuous or not. 
• The integral test is simply a method that helps to determine the convergence or divergence of an infinite series by comparing it to the integral of a related function.
• In simple terms, the integral test states that “if the function f(x) is continuous, positive, and decreasing on the interval [n, ∞) and if the series a _n is also positive and decreasing, then the series converges if and only if the improper integral of f(x) from n to infinity converges.

This can mathematically be expressed as:

Here, it converges if and only if the integral ∫n to ∞ f(x) dx converges.

Read Also: Solving Inequalities

Key Terms: Integral Test, Integral Test For Convergence, Divergence, Sequences, Series, Infinite Series, Calculus, Continuous Function, Relations, Functions

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What is Integral Test?

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The integral test is a method used to determine the convergence or divergence of an infinite series after comparing it to the integral of a related function.

• The integral test is typically based on the concept that a series is the sum of the areas of rectangles under a curve, where each rectangle has a width of one and a height equal to the value of the function at a particular point.
• The integral test states that “if a function f(x) is positive, continuous, and decreasing on the interval [1, infinity), then the infinite series ∑f(n) from n = 1 to infinity converges in case the improper integral ∫1 to infinity of f (x) dx converges.”

Note that if the integral diverges, then the series diverges as well. However, in case the integral converges, the series may converge or diverge. In such cases, additional tests such as the limit comparison test or the ratio test may be necessary to determine convergence or divergence.

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What is Integral Test for Convergence?

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In mathematics, the integral test for convergence is a method which is applied to testing an infinite series of non-negative terms for convergence.

• It was developed by Colin Maclaurin and is also known as the Maclaurin-Cauchy test. 
• The integral test is useful for series where it is difficult to determine convergence or divergence directly.
• It can also be used to establish convergence for series with terms that are related to the values of functions, such as the harmonic series or alternating series.

For integer (non-negative number) N and a function f(x) which is monotonically decreasing, then for the unbounded interval [N, ∞], the function is defined as:

The infinite series   \(\displaystyle\sum_{m=N}^{\infty} f(m)\) converges to a real number if and only if the improper integral \(\int_N^\infty f(t) dt\) is finite. 

Read Also: Operations on Matrices

Integrals Detailed Video Explanation

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Integral Test Conditions

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An integral comparison test is accomplished mostly for the integral terms. Let’s say we have two functions f(x) and g(x) in such a way that g(x) ≥ f(x) on the given interval [c, ∞], then the following conditions should be true.

• If the term \(\int_c^\infty g(x)dx\) converges, then the term \(\int_c^\infty f(x)dx\) also converges.
• If the term \(\int_N^\infty f(x)dx\) diverges, then the term \(\int_N^\infty f(x)dx\) also diverges.

Read Also: Trigonometric Values

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Proof of Integral Test

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The integral test proof relies on the comparison test. It is known that \(\int_N^\infty f(t)dt\) is the sum of the series \(\displaystyle\sum_{m=N}^{\infty} =N=N \int_m ^{m+1} f(t)dt\)

Since, f is a monotonically decreasing function, then f(t) ≤ f(m) for every t in m is ∞ ([m,∞]).

Here, For m>N, \(\int_m ^{m+1}\) f(t) dt ≤ \(\int_m ^{m+1}\)  f(m) dt = f(m)

This means that \(\int_m ^{m+1}\)  f(t) dt ≤ f(m)

As both quantities are non-negative, we can use the comparison test.

If \(\displaystyle\sum_{m}^{\infty}\) = N f(m) converges, then so does \(\displaystyle\sum_{m}^{\infty}\) = N \(\int_m ^{m+1}\) f(t) dt = \(\int_N^\infty\) f(t) dt.

Therefore, it is finite. 

It is just the first step of proof.

Now, let us again assume that the function f is monotonically decreasing; now, for every ‘x’ in [M, m], the equation we get is,

m ≤ f(t) 

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

Hence, the comparison theorem proves that

If \(\displaystyle\sum_{m}^{\infty}\) = N \(\int_{m}^{m-1}\) f(t) dt = f(N) + \(\int_N^\infty\) f(t) dt converges, then \(\displaystyle\sum_{m}^{\infty}\) = N f(m) will also converge, which proves the second part of the theorem.

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Comparison Test

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On the interval [a, ∞], if f(x) ≥ g(x) ≥ 0 then,

• If \(\int_a^\infty\) f(x) dx converges, then \(\int_a^\infty\) g(x) dx also converges.
• If \(\int_a^\infty\) g(x) dx diverges, then \(\int_a^\infty\) f(x) dx also diverges.

While thinking in terms of area, a comparison test makes more sense. If the function f(x) is large than the function g(x), then the area under the function f(x) will be larger than the area under the function g(x).

Therefore, if the area under the larger function \(\int_a^\infty\) f(x) dx converges and is finite, then the area under the smaller function let’s say \(\int_a^\infty\) g(x) dx will also converge and be finite.

Read Also: Integration

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Things to Remember

• There are various tests available for testing convergence but the Integral test is the most useful test for convergence among them.
•  For series with the most irregular sign changes, the integral test is most often useful for testing convergence.
• A series of the form is called p-series where p is constant such that the series is converges if p>1 and diverges if p≤1.
• Since the logarithm has arbitrarily large values, the harmonic series does not have a finite limit; therefore, it is a divergent series.
• In integral test, the initial value where the series starts and the lower limit for the integral must be the same value.
• It doesn’t matter if the first few terms of the series are not positive or decreasing, the test will still work if at some point the terms of the series become positive and decreasing.

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Previous Year Questions

1. Evaluate: \(\int\frac{sin x}{sin 4x} dx \)... 
2. Evaluate: \(\int\frac{x^{3}+x}{x^{4} 9}dx \)… 
3. If a is a positive integer, then the number of values... 
4. The integral ∫cos (logx) dx is equal to… (JEE Main 2019)
5. Evaluate: π4∫0√1−sin2xdx​
6. Evaluate: π2∫π4cos2xdx​
7. Evaluate: ∫sec4/3xcosec8/3xdx​