Comparison Theorem doesn't work, integral of 1/(x^2-1), calculus 2 tutorial

Learn the comparison Theorem for improper integrals, integral of 1/(x^2-1) from 2 to inf. For more calculus 2 tutorials, subscribe to ‪@bprpcalculusbasics‬
Check out my 100 series for more infinite series convergence practice. • 100 series convergence...
----------------------------------------
🛍 Shop my math t-shirt & hoodies: amzn.to/3qBeuw6
💪 Get my math notes by becoming a patron: / blackpenredpen
#blackpenredpen #math #calculus #apcalculus

[@DebrajRakshitIMDeb]

I like the way you give detailed info for each steps. It's really easy to understand this way. Thanks for creating the videos. They are cool!

[@blackpenredpen]

Debraj Rakshit youare welcome

[@MorganMMIX]

U could also compare it to integrating 1/x^k from 2 to inf, where k>1 and 2^k

[@moeberry8226]

Nice video BPRP I just used the limit comparison test to 1/n^2 since this series converges and the limit comparison test yields 1. Then both series also converge and if the series converges then the improper integral also converges.

[@Razib10000]

we can use the limit comparison test for improper integrals!

[@jimschneider799]

Another, perhaps easier way to compute the limit is to use the substitution (x-1)/(x+1) = 1 - 2/(x+1). The limit of the right hand side here is obviously one.

[@louisthurston3067]

As pointed out by others, the limit comparison test should be proved and taught as the primary tool.

[@AlgyCuber]

1:51 that quotation mark looks like a question mark hmm

[@jarikosonen4079]

Comparison theorem looks great then - but it's not needed always. Just dont put sin(x) on the right side this case. Finding the right comparison cases is the problem. How to check that the compared functions are comparable?

[@Utesfan100]

Could we not compare with 2/x^2?

[@MohamedAmin-ct9cx]

By partial fractions = -0.5/(x+1) +0.5/(x-1) so int= ln(sqrt((x-1)/(x+1))) = ln(sqrt(1-(2/(x+1))) substituting by inf and 2 = ln 1 - ln(sqrt(1/3)) = ln ( sqrt (3) )

[@frcai135]

I think you can use the comparison theorem with 1 / (x ^ 1.5)

[@frcai135]

You can prove by induction that x ^2 - 1 > x ^ 1.5 when x is 2 or greater.

[@helloitsme7553]

frcai135 you can just use the fact that 1.5>1 here so it converges

[@pfeffer1729]

Induction is a bad method because it only works on the naturals.

[@elkincampos3804]

Now comparison with 1/((1/2)*x^2)

[@SpeedyMemes]

@@pfeffer1729 I don't think he means n+1 induction, I think he means to show that x^2-1 is increasing faster than x^1.5 and 2^2-1>2^1.5, so its true for larger values also.

[@togrulmaliki4553]

Thank you very much, hocam

[@mihailb0r0koko89]

It’s just easier to say that 1/x^2-1 is equivalent to 1/x^2 as x goes to inf

[@mausamthapa3007]

That was amazing🤩

[@KevinS47]

I decided to give it a go myself before watching the video, and i failed the integral, BUT i got extremely close!! my answer (wrong) seems soo similiar to yours, its crazy! i got: - (ln(2))/3 while you got +(ln(3))/2 very interesting indeed!

[@KevinS47]

I used integration by parts, probably made a mistake at the very beginning, but even if wrong its wired how close it is from the actual answer.. the world of Math! love it!

[@sarikakumari4738]

@@KevinS47 see u can use power series

[@akanshashukla1364]

Thank you❤

[@bsul03420]

Another great blackpenredpen video!

[@blackpenredpen]

Thank you!!!

[@sarikakumari4738]

@@blackpenredpen i got tan^-1(1/2)

[@saarike]

Amazing...

[@boldwintest910]

thanks chief

[@finweman]

I think you could also compare 1/(x - 1)^2 > 1/(x^2 - 1)

[@tomctutor]

∫ (1/(x-1)^2 dx from x=2 to inf = ∫ (1/(x)^2 dx from x= 1 to inf. the latter is a p=2 convergent integral, therefore our problem _∫ (1/(x^2-1) dx from x=2 to inf_ must converge by your stated premise 1/(x - 1)^2 > 1/(x^2 - 1). Well noticed.

[@tomctutor]

@18.Siddhant kumar Not really! The Q was *"does it converge"* ? If you could evaluate it directly then that would be that; this is the proof BPRP actually used. But it is not strictly necessary to evaluate a limit to prove a convergence, in fact Marco has done so using the comparison thm (note I have used limit shifting, for to show that this converges) e.g. try show that ∫ exp(-x^4).dx converges on x from 1 to inf? No need to solve this because you only need to compare with known!

[@tomctutor]

@18.Siddhant kumar Does the power series converge everywhere in the interval? ∫ -[ 1/(1-x^2)].dx = -[ ∫ (1+x^2+x^4+...).dx] = - [x+(1/3)x^3+(1/5)x^5+.. ] only converges for 0⋜x

[@user-uh9bo2im1h]

Okay my attempted using complex analysis int 2 to inf of 1/(x^2-1) dx let x = ix => i*int 2i to infi of 1/(-x^2-1) dx => -i* int 2i to i*inf 1/(x^2+1) dx => -i*arctan x | inf *i u. 2i => congerges

[@joexmorgan]

At 14:08 instead of realizing ln(1) = 0, I used log laws to get that the answer is (1/2)ln(1/3), which is equal to -(1/2)ln(3)

[@gaurabpaudyal]

Thanks bro

[@helloitsme7553]

You could have done that limit much easier, (x-1)/(x+1) = 1 - 2/(x+1) if you plug in infinity in the x you get 1/2ln(1-2/(∞+1)) which is 1/2ln(1) which is 1/2*0 which is 0

[@savanahblaskvitch8253]

1/0 is undefined.. thats why theres a lot more work. you cannot have 0 in the denominator

[@deadfish3789]

It says 0.5*0, not 1/0

[@purewaterruler]

could you use the integral test for series, but in reverse?

[@blackpenredpen]

purewaterruler what do u mean in reverse?

[@purewaterruler]

blackpenredpen well it's usually, in calc classes I've taken, used to test the convergence of a series with an integral. I'm saying to look at the integral, say that since the integrand satisfies the results of the integral test, the integral converges or diverges with sum(2,inf,1/(x^2 -1)). then, for the series, use the limit comparison test(or something else if it works), comparing it with 1/x^2, seeing that the series converges, and concluding that the integral converges. lol at this point I guess I've figured out if works, but I guess it's a nice technique

[@purewaterruler]

blackpenredpen you should be able to use this method for any integrand you can test the convergence of respective series, right?

[@jamesfortune243]

Couldn't you just do a change of variable y = x - 1 and then use the comparison test?

[@Jamelele]

which "arbitols" rule are you talking about? I didn't quite understand what you said sorry

[@blackpenredpen]

Maxstar22 what part?

[@Jamelele]

so you get an answer when you divide 0 by 0, or infinity by infinity

[@yousefal-shinnawi3171]

L'hopital

[@Jamelele]

ahh thank you sorry for having misheard you lol

[@15schaa]

I did something different, and I don't know why it's wrong. 1/(x^2-1) (x^2-1)^-1 ln(x^2-1) ln(inf.^2-1)-ln(2^2-1) ln(inf.-1)-ln(4-1) ln(inf)-ln(3) inf.-ln(3) Diverges.

[@cristicris9975]

It's wrong because you had to have (2x)/(x^2-1) so then it will be ln(x^2-1) since you don't have that you can't say 1/(x^2-1) = ln(x^2-1)... that is a compound function and in integration to use the normal formula with a compound function you have to have product of the function and it's derivative say f(x) = x^2-1 if you had 1/f(x) * f'(x) dx then yes, the answer will be ln(f(x)).

[@shivanshsharma6921]

Only linear eqn like( ax+b ) can be integrated if it's inside another function like ln (a x + b) or tan inverse (a x + b ) for functions like (aX square + bx + c) inside another function you can only integrate by converting it into simple or linear equation

[@Treegrower]

I've always wondered why you have to write improper integrals as limits. Why can't we just write it the easy way? Wouldn't we get the same answer anyways?

[@manikin3095]

You have to write it as a limit because infinity is a concept, not a number. Formally, you can't say "infinity - 1" because again, infinity is not a number. However, you can say lim as t -> inf. (t - 1) because then, formally, you can get an answer TLDR: yes you get the same results, but it's for formality

[@ToneWarzMusic]

The reason why you take the limit is because "plugging in" infinity does not always give a defined result, like the indeterminate form shown in the video. For example, the limit as x -> inf of xe^(-x) will approach 0. However, "plugging in" infinity will give you infinity times 0, which is indeterminate.

[@AnkhArcRod]

Title needs to be fixed. Since you are teaching math, you need to be precise. In this case, the correct statement is one comparison theorem test did not work...or better still, just find the correct comparison to work with. Int (1/(x^2-x)) is always greater than Int (1/(x^2-1)) for interval [2,inf) and the former integrates to ln 2. Hence, given integral is convergent.

[@matikb.6733]

but isn't 1/(x^2-1) = 1/((x+1)(x-1)) = (1/(x+1))*(1/(x-1))

[@youmah25]

he used the partial fraction expansion of 1/(x^2-1) i hope this helps

[@jamesstrickland833]

He converted it from multiplication to addition to make the integral easier to work with

[@MultiOmps]

How come infinity - infinity can't be assumed as 0. Isn't any number subtracted from itself zero.

[@egybizonyosember]

Infinity is not a number, but a limit. If, for instance, the limit of a function f(x) is infinity as x goes to infinity, it just means that you can get a ginormous number out of the function by putting a ginormous number inside of it. Likewise, if the limit of a function at 2 is, say, 5, it means that you can get as close to 5 as you wish by putting a number very close to 2 inside the function. As an example, consider the function f(x) = 7^x-3^x. I believe it is obvious that its limit in infinity is infinity. Yet, if you separate it into the difference of two limits, you get infinity - infinity. This is the reason why this kind of limit is called indeterminate form. There are others as well.

[@MultiOmps]

However, though can't infinity be thought of as the final number in the real number line?

[@egybizonyosember]

No. First of all, there is no such thing as a final number in the real line. You can tell me any number, and I'll always be able to say one that's bigger. Secondly, as I have tried to stress in my first comment, that is not how is is treated, and that's why infinity - infinity is not always zero. It doesn't mean a concrete number, it means "And arbitrarily big number subtracted from another arbitrarily big number". With all due respect: I feel like you're missing some key understanding here, and I'm not willing to give you an intro course to calculus. Check out Coursera, Khan Academy(especially this one), Edx or Udemy for great courses that will give you an understanding that I cannot. There are many free courses.

[@MultiOmps]

I'll check these courses out then.

[@willyou2199]

ln(infinity) = infinity. if infinity is treated as a number, then infinity would be a solution to ln(x)=x, which there are no solutions (except in the complex plane.) Not all infinities are the same size, here, the infinity inside the ln is necessarily larger than the infinity in the rhs, but they're both so incomprehensibly big

[@cheriekorer352]

T

[@dalek1099]

The comparison theorem works just fine thats why I've put a thumbs down on this video and this problem is a very standard sort of comparison theorem question that could be asked at University. Firstly, when you use the comparison theorem you can't just choose what integral you want to compare it with you have to derive the relevant inequality for the problem at hand. Here is how you would do it since x>=2 x^2>=4 x^2/2>=2 therefore x^2/2>=1 x^2/2=x^2-x^2/2

[@Omikron35]

why don't you juste say that 1/x²-1 ~ 1/x² whose integral converges ? what you're doing is so complicated for nothing really isnt it ?

[@ashspawk1206]

actually you're all wrong...2 + 2 is 4 QUIK MAFFS

[@atharvas4399]

you could compare it with 1/x^3

[@finweman]

Don't think so. 1/x^3 < 1/(x^2 - 1) for x >= 2.