Dirichlet Test & Absolute Convergence for Improper Integrals Type 1

Convert to note

The Dirichlet test provides a method to determine the convergence of improper integrals of type 1, particularly for products of functions. It is especially useful when one function has unbounded but uniformly bounded integrals, and the other is monotone decreasing to zero.

Conditions for the Dirichlet Test

Consider two functions f and g defined on [a, ∞) → R, such that:

f is integrable on every interval [a, B] for B > a.
The integrals of f are uniformly bounded: there exists a constant C > 0 such that |∫ₐᴮ f(x) dx| ≤ C for all B > a. This bound C is independent of B.
g is monotone (either monotonically increasing or decreasing) and bounded.
lim_{x→∞} g(x) = 0.

If these conditions are satisfied, then the integral ∫ₐ^∞ f(x)g(x) dx converges.

Application Examples

Example 1: ∫₁^∞ (sin x / x^p) dx for p > 0

Let f(x) = sin x and g(x) = 1 / x^p.
g(x) is monotonically decreasing to 0 as x → ∞ for p > 0.
The integral of f is uniformly bounded: |∫₁ᴮ sin x dx| = |cos 1 - cos B| ≤ 2.
Therefore, by the Dirichlet test, ∫₁^∞ (sin x / x^p) dx converges for all p > 0.

Example 2: ∫₀^∞ (sin x / x) · e^{-ax} dx for a > 0

Here a is a positive constant. The integral is split at x = 1 for analysis:

∫₀¹ (sin x / x) e^{-ax} dx: This is a proper integral because the integrand has no singularity at x = 0 (lim_{x→0} sin x / x = 1). It converges.
∫₁^∞ (sin x / x) e^{-ax} dx: Apply the Dirichlet test.
- Let f(x) = sin x / x. Its integral is uniformly bounded: |∫₁ᴮ (sin x / x) dx| ≤ |∫₁ᴮ sin x dx| = |cos 1 - cos B| ≤ 2.
- Let g(x) = e^{-ax}, which is monotone and lim_{x→∞} e^{-ax} = 0.
- Therefore, ∫₁^∞ (sin x / x) e^{-ax} dx converges.
Since both integrals converge, the original integral converges. For a deeper understanding of this technique, see Using the Residue Theorem to Evaluate Improper Integrals.

Absolute vs. Conditional Convergence

An integral ∫_a^∞ f(x) dx is said to:

Converge absolutely if ∫_a^∞ |f(x)| dx converges.
Converge conditionally if ∫_a^∞ f(x) dx converges, but ∫_a^∞ |f(x)| dx diverges.

A key fact is that absolute convergence implies convergence, but the converse is not true.

Example: ∫₀^∞ (sin x / x) dx

Convergence: The integral converges (e.g., by the Dirichlet test as in Example 1 with p = 1).
Non-absolute convergence: To show ∫₀^∞ |sin x / x| dx diverges, break it into intervals of length π:

∫₀^∞ (|sin x| / x) dx = Σ_{n=0}^∞ ∫_{nπ}^{(n+1)π} (|sin x| / x) dx

Substituting x = nπ + y gives:

= Σ_{n=0}^∞ ∫₀^π (sin y / (nπ + y)) dy

Since y ≤ π, nπ + y ≤ nπ + π = (n+1)π. Therefore, 1/(nπ + y) ≥ 1/((n+1)π). This gives the lower bound:

≥ Σ_{n=0}^∞ (1 / ((n+1)π)) ∫₀^π sin y dy

= (1/π) Σ_{n=0}^∞ (1 / (n+1)) · 2

= (2/π) Σ_{n=1}^∞ (1/n) = ∞

Since the harmonic series diverges, the integral of |sin x / x| diverges. Hence, ∫₀^∞ (sin x / x) dx is conditionally convergent. This type of integral can also be approached using Computing Improper Fourier Integrals Using Complex Analysis Techniques.

Handling Integrals with Lower Limit -∞

Improper integrals of the form ∫{-∞}^a f(x) dx can be transformed into the standard form by a substitution like x = -t. This yields ∫{-∞}^a f(x) dx = -∫_{-a}^∞ f(-t) dt, which can be analyzed using the same tests. When working with such transformations, the Understanding Cauchy’s Theorem and Complex Integrals Explained can provide additional foundational insight.

[Music] so welcome to the lectures on engineering mathematics one and this is

lecture number 23 we are talking about the integral calculus in particular improper integrals so we will continue

the discussion on the convergence of type one integrals now we have one more test which was not discussed in the

previous lecture that is called the distillate test and this is again a very important test for testing the

convergence of improper integrals of type one so here we suppose that this F and G these are the two functions

defined from this a to infinity and they are taking the real value and they are such that that F is integrable on each

interval so a B and B can take any value greater than a so this f is integrable on each integral A to B and the other

one the integral A to B this FX this is the second condition these integrals are uniformly bounded so integrals means

that this B can take any value so we are not restricting on B so these integrals here are uniformly bounded what does

that mean that there exist number C here the constant C greater than 0 such that the value of this integral the absolute

value of this integral a to b fxdx is less than equal to that constant and for all B greater than a naturally be some

finite number we are talking about so here for any value B here which is greater than a if this integral is

bounded by a constant C it is the C is not depending on the number B here so in that case we call that this is uniformly

bounded so this bound is independent of this B then we call that this integral is uniformly these integrals are

uniformly bounded so these are the two conditions on F F is integrable and this is uniform these are these integrals are

uniformly bounded and the function G is monotone and bounded so again that either monotonically increasing or

monotonically decreasing and they bounded so the values of dysfunctions are also finitely bounded and then we

have this limit as X approaches to infinity this GX is zero so this is a G as a monotonic function which is

approaching to zero as X approaching to infinity so that is another condition and in that case this improper integral

e a to infinity FX GX DX so we have now the product here in the integrand FX GX and this converges that's a result so we

will not go through the proof of this result so what is important here the important is to check that these

integrals here they are uniformly bounded so meaning we have to compute this integral and see whether we can

bound that integral by some constant C which is independent of this number be here for any finite number B if we can

do that so we have this condition uniformly boundedness and the second one then G is monotone and so monotonically

it is going to 0 as X approaching to infinity so in that case so these are the two main conditions here and that

those conditions we have that this integral the product here converges so this is a very useful integral because

if we know that one of them so for example G is here this bounded and monotone going to 0 here and then these

integral over F converges so then we can conclude about this product as well that what will happen to this integral and it

will converge if these conditions are satisfied so here let's go through this example where we can use this result

easily to prove such integrals 1 to infinity sine x over X power P is convergent for P greater than equal to 0

so for any P here greater than 0 this integral 1 to infinity sine x over X power P is convergent we will show with

the help of this de rich let test so in this case we let that FX is equal to sine X so we have this function

here FX s sine X and the other one GX we will take 1 over X power P so how this is useful now because we know about this

G that this is monotonically decreasing to zero as X approaches to infinity for this P+ and this integral of the sine X

which we can easily show that one to be the sine X DX so this integral will be there - this cos x and then we have this

limit A to B so which will be this cos one minus the COS be this integral value and then the absolute value of this we

have to consider so this course is bounded by one always and we can use this inequality here that this is less

than cos one plus cos B and in that case this will be bounded by two so we can easily show that this integral here is

is bounded by by two for any value of B we take which is greater than 1 less than infinity so what we have seen that

there is a uniform bound here so whatever values of B we take the value of this integral will be bounded by two

and this is what we we want for uniform boundedness of this integral now the other function G which is a

monotonically decreasing function and naturally this tends to 0 as X tends to infinity for any value of P positive so

using that there is less test now so we can talk about this product the integral of this product sine X and product with

1 over X power P that means this integral sine x over X power P DX from 1 to infinity this converges for any value

of P greater than 0 so this distance test is very useful now that we have just considered as a ratio of the two

function again so one was FX and another GX and this function FX had this nice property about this uniform

integrability and the second one was monotonically decreasing to zero therefore we could conclude with that

the wristlet test that this converges another half similar kind we can test the convergence of this zero to infinity

sine x over X power minus X this integral here so for any value of this a positive so what we will do we can prove

this again easily so this is e power minus x cos x alright so this was the first problem if power so sine x over X

E power minus X so we have already done before that was sine x over X power P when 1 to infinity and now we will do

with sine x over X multiplied by E power minus X so in this case we will again this use the idea that we can break this

easily 0 to 1 and then we have sine x over X E power minus X DX Plus this 1 to infinity and we have sine x over this X

and E power minus X DX this function so the first integral is a proper integral can proper integral the reason is clear

because the sine x over X has the value 1 as X approaches to infinity and this is 1 so the limit here of the integrand

exists when X approaches to 0 and at all other point there is no no singularities there is no unboundedness so this is a

proper integral the first one the second one now because of this X infinity we have to test that what type of result we

get from here so now we take this integral here a to be so note that the integral 1 and then above 1 we can take

this B is sine x over X DX so if we take this integral here sine x over X DX a 1 this X is going from 1 to any number

this B so here the X we can replace because this is the minimum value if we replace this one so that the integral

will be the larger one so we have 1/2 B and this X is replaced by one the lowest possible values so that the integrand is

the larger one so we have this integral bigger than this one as sine X DX and this we know now that we can integrate

this with minus cos x and then 1/2 B so this value will be cos 1 and minus D cos B and we can take that absolute value

there so this absolute value of these functions will be again the absolute value here which is bounded by 2 so we

can prove that that this integral here sine x over X the sine x over x want to be for any value of b greater than 1 any

finite value this is uniformly bounded and this value is 2 so this is uniformly this integral is uniformly bounded by 2

and this exponential minus X function so exponential minus X function is is monotone and it approaches 2 to 0 if we

take the limit as X approaches to infinity this e power minus x that approaches to 0 so we have that property

also satisfied so then we can use this directly test there because the 1 integral with the sine x over X we have

seen that that is monitor that is a uniformly bounded and the other one here the GX which is e power minus X it is a

it is going to 0 as X approaches to infinity so in this case we can apply now the dirichlet results do reach let's

test which says that this integral converges so this integral converges and the other one is proper integral so

we have the original integral here zero to infinity sine x over X that converges well so we do one more problem here that

is problem number two so here we will it's a similar kind of problem we have 1 minus e power minus x cos x over X

square in this case and the idea again we can break into two parts so I will take your a to infinity and then this we

have 1 over X square cos X as one integral so we have cos x over X square DX and the other one is a to infinity

and we have E power minus X and then cos x over X is Square D X so this first integral this is converges absolutely

this converges converges absolutely and the reason is clear because by the comparison test we can easily conclude

because this cos x over over x square this is less than this 1 over X square indeed this absolutely concept we will

we will explain again so here the absolute only means that the if we take even the absolute value of the integrand

this is this converges so this cos x over X square is bound is less than equal to 1 over X square because it's

cos x the values are bounded by 1 so we have 1 over X square and we know this integral as a to infinity 1 over X

square that converges and in that case this integral will also converge because this is dominating function 1 over X

square and the whose integral converges so naturally that integral will converge the second one here we can again use the

same argument because again this e power minus x cos x over this x square so this value

here we'll be bounded by by one because II power this is minus X is also decreasing function so it will take the

highest value when we have then X will approach to this a so in any case this will be also bounded by one so this

second integrand is also bounded by 1 over X square and by the comparison test we can again conclude that this also

converges absolutely we can also use here the drizzler test because we can take this e power minus x so if you want

to use this distillate test here we can take this function e power minus x over x square and this function approaches 2

to 0 as x approaches to infinity as X approaches to infinity this is monotone function e power minus x is decreasing

in 1 over X square is also decreasing so both the functions so here we have the decreasing function for whatever values

of x and this cos x the integral of this a to infinity cos x we can again show that this is bounded by two so we can

use that drusilla test like we have done in the earlier problem so this is similar to the problem number one so we

can here also show that this converges so this also converges and the given integral they are 1 minus e power minus

x cos x over x square DX converges so now just note here that this integral of this type minus infinity to please FX DX

we are not you normally discussing here we are just discussing the integrals where we have infinity in place of this

v so this a to infinity type of integrals we are discussing but this also we can easily deal because we can

substitute for example here x is equal to minus t and then what will happen that this integral will be just minus p

to infinity so again we have a similar type integral which we are discussing for the convergence where the infinity

appears above so if we have this type of integrals we can just by simple substitution we can convert

into this type of integral and then discuss the convergence the way we have discussed earlier so this is the point

here we were talking about the absolute convergence another important factor here that what is the absolute

convergence so as I said before the absolute convergence means that this integral converges absolutely meaning

that if with the absolute value over the integrand if this converges so if this integral converges we call that this

integral converges absolutely because we have taken the positive values of this integrand for the range here zero to

infinity so if this converges then we call that this integral converges absolutely and there is a one more term

which is used here that this integral converges conditionally meaning that this integral converges but it does not

converge absolutely so here meaning is that this converges but not absolutely in that case we call

usually that this integral converges conditionally so here like coming to this example here 1 to infinity sine x

over X power P DX and we can show now that this integral converges absolutely for P greater than 1 for any value of P

greater than 1 this integral converges absolutely the idea is clear which was also mentioned in previous example a bit

so here we have this in T if we take the absolute value of this integrand that means the sine X and X power P so here X

are taking positive values and P is greater than now 1 so here we have this sine X absolute value over X power P

because this is positive so we don't have to use the absolute value here and this is bounded by 1 over X power P

because the sine X will never take value greater than 1 so we can bound this by 1 over X power P and we know now by the

comparison that the integral 1 to infinity 1 over X power P this integral converges so what

we have shown here that this integral also converges because this we have by the comparison test here so we have

taken the integrand here that the the sine x over X power P which was less than equal to 1 over X power P and we

have we know the result that this converges the integral of 1 over X power P 1 to infinity this converges and

therefore by the comparison test we have shown that this converges meaning that this integral in other words converges

absolutely yeah so we can take and see that because at person this is taking positive negative values in the range so

if we take indeed all the positive values in that case also this integral converges but this is important that we

have shown here for P greater than 1 that this integral converges absolutely when P is greater than 1 we will see

later on that when P is equal to 1 this integral does not converge absolutely indeed it converges but not absolutely

so going to the next problem so this is there is a result also that this is such integral converges if this converges

meaning with the absolute value because this is going to be a larger value than this value of the integral so if this

converges naturally this converges because here we have positive negative and many this cancellation will come and

this value will be less than the value here with while taking the absolute value of the integral so this converges

if this converges but the converse is not true meaning that so this will converge if this converges but if this

converges this may not converge yeah so if the integral converges the integral may not converge absolutely and that's

the way standard exam which you will discuss now here that 0 to infinity the sine x over X DX this

integral converges conditionally meaning that this integral converges but when we take the absolute value here over the

integrand then this integral does not converge so first we will show that this integral converges which is a trivial

task now so 0 to infinity sine x over DX we have written as the sum of these two integral 0 to 1 sine x over X DX and 1

to infinity sine x over X DX and note that the first integral is is a proper integral so naturally it converges and

the second one we have this infinity this we have to to discuss so this is a proper integral and here we have seen in

this example 1 that year even X power P and P is greater than 1 so that integral converges so naturally this integral

also converges sine x over X we have seen already in example 1 so that means that this integral converges so the

integral 0 to infinity sine x over X DX converges and next we will show that when we take the absolute value over the

integrand that integral does not converge so we have this example which says that the converse is not true

because here we have the convergence of the integral but the integral does not converge absolutely the other way around

this is obvious because if we're taking all the positive values is still we can get this integral so naturally when we

take the mm when the function takes positive and negative values it will certainly converge so here we will show

now that this integral does not converge and this is the bit involved now so we take this absolute value sign x over X

DX and we write down this as a sum of n goes sum of over this n goes from 0 to infinity n pi and n plus 1 pi and sine X

absolute value over X DX so here this integral is is exactly this integral so we have broken here like n is equal to 0

so we are going from 0 to PI and then this integrand and the value then we are going for n is equal to 1 so that means

PI to 2 pi and this integrand and so on so we have taken the 0 2 pi pi to 2 pi and then 2 PI to 3 PI and so on so this

integral 0 to infinity we have broken into several these small segments over the range so 0 2 pi PI to 2 pi 2 pi to 3

PI and so on which we have written in this summation form so this is the integral and now we will make a

substitution here that X is equal to n PI plus y so we make the substitution X is equal to n PI plus y in this integral

so by substituting this so DX will become this dy and then this n PI will be replaced accordingly and n plus 1 PI

as well so when we substitute this in this integral so when the X was taking n PI values the Y will take 0 value so

again to see here because this relation was X is equal to n PI plus 1 so the X was taking the lower value as n PI so n

PI plus y so this Y will take as the value 0 so this goes from 0 now and when X was taking the value and PI plus this

PI so in that case we have this n PI plus y so the Y will take the value PI so our integral is now 0 to PI and this

absolute value sign and this X is n PI plus y so we have n PI plus y and here also we have substituted for X that is n

PI plus y so we will make another use of this inequality a of this equality that Sen

sine n pi plus y sine n pi plus y is equal to -1 power n and sine Y because for example n is 1 so we have sine pi

plus y which is minus sine Y and similarly for other ends so we have this equality the techno Matic equality which

we will use here now in this numerator because we have sine n pi plus y so Y using this we have now in the numerator

minus 1 power N and then we have this sine Y over n pi Y dy the same integral which was here so this sine n PI plus y

is replaced by -1 power n sine Y so now this summation as it is we have 0 to PI and this minus 1 power n because we are

talking about the absolute value so that will be as it is and the sine Y again so here we should note that we have we are

not using now the absolute value because the sine Y is will take positive values from 0 to PI it takes value 0 there and

then at PI by 2 1 and then it gradually goes to 0 so this never takes negative values in 0 to PI therefore we have

removed this absolute value here so divided by sin PI plus y and dy so in this case now this denominator if we

take a look here so Y in this case for this integral is taking zero value through the pi value so if we want to

make this integral larger than some other integral here so we can replace this Y by the larger value or this

denominator will be the taking the largest value here at pi here so we have n pi plus pi so this Y is replaced by

this pi so that this is this integral here becomes a smaller integral for this integral and therefore we have this

equality that this integral is larger because this is taking now the integrand is taking lower value because this

denominator was replaced by the lowest well the highest value here which is pi there so n pi plus pi so this is a

smaller integral and now here this is nothing but because there is no y here so we can take this out so we have n

sorry despite we can take out so we have pi there and the integral also we can evaluate so we have this summation and

go 0 to infinity and 1 over n will remain as it is so we have also this pi here so 1 over pi and this n plus 1 so 1

over pi and n plus 1 or again let me write it down so we have this summation and goes from 0 to infinity and this is

1 over pi and n plus 1 and integral 0 to pi and we have this sine Y dy so this will be minus cos Y and then 0 to PI so

when we substitute pi we will have again minus 1 there so minus minus will be plus and then minus minus will be plus

again so once so we have to value to here so this is replaced by 2 so we have the value 2 over pi and the summation

over 1 over n plus 1 so this integral is equal to this one and this is a well-known series here which is a

divergent series so that means the sum is infinity here so what we have seen that this integral which we have started

with the absolute value this is greater than this sum of this series which is infinity so that means this integral is

a divergent integral because the value of this integral will be larger than the value of this sum which is coming

already to infinity so in that case we have that this improper integral the given integral

this diverges so the conclusion we have the desolate test which says that if we can show this uniform boundedness of

this integral for all be greater than one and G is a monotone decreasing to zero as X approaches to infinity then we

have this product in the integral a to infinity this DX converges and we have also discussed about the absolute

convergence there and we have seen this nice example which does not converge absolutely but it converges if we don't

take this absolute value for the integrand so these are the references used for preparing these lectures and

thank you very much

What is the Dirichlet test and when should it be used?

What are the four key conditions needed for the Dirichlet test to apply?

Does the Dirichlet test apply to ∫₁^∞ (sin x / x^p) dx for all p > 0?

What is the difference between absolute and conditional convergence for improper integrals?

How can we prove that ∫₀^∞ |sin x / x| dx diverges?

How do you handle improper integrals with a lower limit of -∞?

Can the Dirichlet test be applied to integrals like ∫₀^∞ (sin x / x) e^{-ax} dx for a > 0?

Keep this summary

Save it to LunaNotes and it becomes a real note in your library — editable, searchable, and ready to turn into flashcards or a diagram. Free to start.

Save to LunaNotes

Or summarise for another video.

32K+ people have turned 65K+ videos into notes

“Very useful tool. I'm using it for the last couple weeks and I'm loving it so far. Makes learning on YouTube a great experience.”

Bhai Chrome Web Store review

This summary and transcript were automatically generated using AI with the Free YouTube transcript summary tool by LunaNotes.

Related summaries

Using the Residue Theorem to Evaluate Improper Integrals

This lecture explains how to apply the residue theorem from complex analysis to compute improper integrals over infinite intervals, including a detailed example involving rational functions and the Cauchy principal value. Key takeaways include conditions for convergence, the role of even functions, and a step-by-step residue calculation method.

Computing Improper Fourier Integrals Using Complex Analysis Techniques

Learn how to evaluate improper integrals involving sine and cosine from negative to positive infinity using complex contour integration, residue theorem, and Jordan's lemma. This guide explains replacing trigonometric terms with complex exponentials, identifying poles, and applying these theorems to compute Fourier integrals with detailed example solutions.

How to Integrate Complex Logarithm Functions Using Branch Cuts

This lesson explains step-by-step how to integrate the complex logarithm function \( \ln x / (1+x^2) \) from 0 to infinity using contour integration and branch cuts. Learn how to set up the integral in the complex plane, apply the residue theorem, and use the ML inequality for evaluating integrals over semicircular contours.

Understanding the ML Inequality: Bounding Contour Integrals in Complex Analysis

This video explains the ML inequality (estimation lemma), a fundamental theorem in complex analysis that provides an upper bound on the magnitude of contour integrals of piecewise continuous complex functions. It details the theorem's conditions, provides a step-by-step proof using a key lemma, and illustrates the concept with a geometric example involving an arc of a circle.

Using the Residue Theorem to Evaluate Definite Integrals Involving Sine and Cosine

This video lesson demonstrates how to apply the residue theorem from complex analysis to evaluate definite integrals involving sine and cosine functions. It explains transforming trigonometric integrals into contour integrals on the unit circle using complex exponentials, then finding residues at singularities to compute the integrals efficiently, including examples with detailed calculations.

Start taking better notes today with LunaNotes

Start for free