CSIR NET MATHEMATICS| Real Analysis | Theory and Solution of Improper Integral Problem JUNE 2018

In this video, we are going to discuss the problem of real analysis from CS Ionate, June 2018. Example number 66 from booklet A which is of marks 4.75. The problem is let alpha equal to integration 0 to infinity 1 upon 1 plus t square dt. Which of the following are true? The first option d alpha by dt equal to 1 upon 1 plus t square. The second option is alpha is 1. The third option is log of alpha equal to 1 and the fourth option is sine alpha equal to 1. Now here in the given problem alpha is equal to integration 0 to infinity. So one of the limit is infinity so it is a improper integral and more than that it is a first-kind improper integral. So here first we are going to understand the evaluation process of first-kind improper integral. So first case. Where alpha equal to integration 0 to infinity f of t dt here the upper limit is infinity. Now to evaluate this integration first we convert into the proper integral by applying the limit. Limit t tends to infinity integration 0 to t f of t dt here integration 0 to t is proper integral and we know how to evaluate the proper integral after the evaluation of proper integral We apply the limit t tends to infinity. Case 2 where alpha equal to minus infinity to 0 f of t dt. Now the similar way limit t tends to infinity minus infinity t to 0 f of t dt. The third case is a combination of previous two case that means alpha is equal to integration minus infinity to plus infinity. So both the limits are infinity f of t dt. So first we split this integration into two parts minus infinity to c Where C is a any finite fixed number. f of t dt plus c to infinity f of t dt. So it's a combination of first previous two case and we can write this limit limit t tends to minus infinity integration t to c f of t dt plus limit t tends to positive infinity integration c to t f of t dt. Now the given problem is a case one where upper limit is a infinity. So we are going to evaluate the given integration So, integration 0 to infinity 1 upon 1 plus t square dt. Now, rewrite this integration in the form of proper integral with the limit. So, limit t tends to infinity, integration 0 to t, 1 upon 1 plus t square. Now, integration of 1 upon 1 plus t square is tan inverse t, evaluate from 0 to t. So, the upper limit is capital T, tan inverse capital T minus tan inverse 0. The value of tan inverse 0 is 0. So, limit t tends to infinity, tan inverse. capital T. If T goes to infinity, tan inverse T goes to pi by 2. Therefore, alpha has value pi by 2. Now, we are going to check all the cases. Alpha equal to pi by 2, which is constant. So, its derivative is 0. So, the first option is d alpha by dt equal to 1 upon 1 plus t square, which is wrong. Now, here pi is irrational. So, pi by 2 is also irrational. Therefore, For the second case. Alpha is Rational which is also a wrong. Now third option is log of alpha. So log of alpha equal to log of pi by 2 which has value 0.196. Therefore the third option is also wrong. Now sine alpha is equal to sine pi by 2 and psi pi by 2 has value 1 which is the fourth option. Therefore fourth option is correct. So this problem has only fourth option is correct. Thanks for watching. Like, share, subscribe and press the bell icon of NextGenMass.