Improper Integral of type 1

In this video, I showed how to rewrite an improper integral of type1 to make iteasier to compute

[@acerovalderas]

You are great. I am lucky to have found you.

[@savannahkim6679]

Love the flow. You are doing great. Keep on!!!!

[@learning-iz-fun]

Limit can be solved by observation as well. In the product of linear and exponential term, the exponential will always grow or shrink faster than linear

[@iithomepatnamanojsir]

Very nice sir

[@josegerardodias5591]

Very nice! You are great!

[@semanurgulhan9520]

You're amazing thank you♥️

[@RanBlakePiano]

This is terrific !

[@punditgi]

The proper way to learn about improper integrals is to watch Prime Newtons! 🎉😊

[@ZolileZicwele-lb6bq]

Yeah, thats proper

[@AzmiTabish]

'Definite'ly 🙂

[@amiahooman]

Nice like begging

[@arbenkellici3808]

Hello professor! I need your help to solving: lim x approaches 9- (9-x) 1/4=0 by delta epsilon proof Thank you!

[@henrikljungstrand2036]

It seems you would need to change the x on the lhs of your expression into x+delta, change the 0 on the rhs into y+epsilon (y being an implicit function of x), and then assume you get such an arbitrary epsilon as an input, and trying to find/make up a suitable implicit function to compute delta as an expression of epsilon; then you would need to compute two different limits, one where epsilon is assumed positive and one where epsilon is assumed negative (where you may compute delta out of epsilon in different ways if you like); and in both cases showing that the expression in x and delta is behaving in such a way that computing it and then taking the limit as epsilon converges to 0 makes your delta also converge to 0, and then forces y to be 0 in both cases. Note that y is not allowed to be dependent on epsilon (or on delta, since delta is dependent on epsilon), only on x, while delta is allowed to depend on both x and epsilon. I have given you the outline for the general theoretical reasoning, now go and do the practical work yourself!

[@bastianmora4780]

1

[@holyshit922]

Gamma function ? By very simple substitutoion we will get Gamma function

[@Phantom_Reaper_]

That's a cool approach, but i feel like it's using a higher level concept that most people won't be aware of at the level of class this question would be given

[@holyshit922]

@@Phantom_Reaper_I have used Gamma function for expressing product lately which I have got from ODE so I thought about it but I agree that it is better to use integration by parts and limits unless it is some strange introduction to the Gamma function

[@Phantom_Reaper_]

@@holyshit922 Ive only seen a little bit about the gamma function off of KZbin, but does this integral just become - Γ(0) when you substitute t = -x? That's quite an elegant solution even though it may be unnecessary at this level

[@holyshit922]

@@Phantom_Reaper_ After substitution you will get -Γ(2) -te^{-t} = - t^{2-1}e^{-t} and Γ(2) = 1 Γ(z) has singularity at zero I tried to calculate coefficients of Chebyshev polynomial via power series solution of ode I derived equation in following way I know that T_{n}(x) = cos(n*arccos(x)) Let t = arccos(x) y(t) = cos(nt) Let's differentiate y(t) twice y''(t) = -n^2cos(nt) y''(t) = -n^2y(t) y''(t) + n^2y(t) = 0 Now I changed independent variable t = arccos(x) in this equation and I got (1-x^2)y''(x) -xy'(x) + n^2y(x) = 0 Now I tried to solve it using power series y(x) = \sum\limits_{m=0}^{\infty}c_{m}x^{m} While solving this equation I have got following recursion (m+2)(m+1)c_{m+2} - (m-n)(m+n)c_{m} = 0 and solved it for c_{m+2} (Because we expecting polynomial it is possible to solve also for c_{m}) c_{m+2}= \frac{(m-n)(m+n)}{(m+2)(m+1)}c_{m} c_{m} can be expressed in terms of product \frac{1}{m!}\cdot\prod\limits_{k=1}^{\lfloor\frac{m}{2} floor} (m - 2k - n)(m-2k + n) And I have got Γ(0) after expressing this product \prod\limits_{k=1}^{\lfloor\frac{m}{2} floor} (m - 2k - n)(m-2k + n) in terms of Gamma function but only for n=m=0

[@kevinmadden1645]

​@@Phantom_Reaper_Using an A-Bomb to swat a fly .

[@JSSTyger]

I'll say -1.

[@Ivan-fc9tp4fh4d]

I think the last sentence was not correct. Limit does not converge to "-1". Limit IS equal to "-1".

[@PrimeNewtons]

I actually meant to say the Integral converges to -1.

[@arbenkellici3808]

Sorry (9-x)^1/4

[@henrikljungstrand2036]

It would be better to redact your original comment.