 In this video, we are going to discuss beta function part 3. Now learning outcome, at the end of this session, students will be able to evaluate definite integrals by beta function. Now in this video, we are going to discuss beta function as improper integral. And it is defined as beta of mn is equal to integration from 0 to infinity x raised to m minus 1 divided by 1 plus x raised to m plus n dx. And now we will see how we will get this formula. This is a third form of beta function. We have already seen the two different forms of beta function. Now let x is equal to tan square theta. Then you get dx is equal to 2 tan theta, 6 square theta, d theta. Now accordingly you can change the limit when x is equal to 0 means tan square theta equal to 0 means theta equal to 0. And when x tends to infinity, you get theta equal to pi by 2. Therefore, we can write integration from 0 to infinity x raised to m minus 1 divided by 1 plus x raised to m plus n dx is equal to integration from 0 to pi by 2. Now putting x equal to tan square theta, we will write tan square theta raised to m minus 1 divided by 1 plus tan square theta raised to m plus n and the value of dx is 2 tan theta 6 square theta d theta. Now which is equal to taking two outside that it twice 0 to pi by 2 that is tan square theta raised to m minus 1 into tan theta can be written as tan theta raised to 2 m minus 1 into 6 square theta d theta divided by 1 plus tan square theta is 6 square theta. We can write 6 square theta raised to m plus n which is equal to twice integration from 0 to pi by 2. Now tan theta raised to 2 m minus 1 as it is tan theta raised to 2 m minus 1 and 6 square theta raised to m plus n and 6 square theta we can write 6 theta raised to 2 m plus 2 n minus 2 d theta. Therefore, integration from 0 to infinity x raised to m minus 1 upon 1 plus x raised to m plus n dx is equal to twice. Now writing tan in terms of sin head cosine you get twice 0 to pi by 2 sin theta raised to 2 m minus 1 upon cos theta raised to 2 m minus 1 into cos theta raised to 2 m plus 2 n minus 2 d theta that is 1 by 6 theta means cos theta which is equal to we can write twice 0 to pi by 2 sin theta raised to 2 m minus 1 and simplify these two we get cos theta raised to 2 n minus 1 d theta which is the trigonometric form of beta function or second form of beta function which is equal to we can write beta of m n. Therefore, beta of m n equal to we can write integration from 0 to infinity x raised to m minus 1 upon 1 plus x raised to m plus n dx. Now we will see the example evaluate the integral integration from 0 to infinity x raised to 6 into bracket 1 minus x raised to 10 bracket complete whole divided by 1 plus x raised to 24 dx. Now we can write integration from 0 to infinity x raised to 6 into bracket 1 minus x raised to 10 divided by 1 plus x raised to 24 dx is equal to you know here multiplying x raised to 6 inside we get integration from 0 to infinity x raised to 6 divided by 1 plus x raised to 24 dx minus that is x raised to 6 into x raised to 10 that is x raised to 16 divided by 1 plus x raised to 24 dx. Now these two integrals are in the form of integration from 0 to infinity x raised to m minus 1 divided by 1 plus x raised to m plus n. Now here m minus 1 is 6 which you can write first integral can be written as which is equal to integration from 0 to infinity x raised to 6 we can write x raised to 7 minus 1 that is here m is 7 and m plus n is 24 therefore we get n as 17 that we can write 1 plus x raised to 7 plus 17 that is m plus n here m is 7 and n is 17 and m plus n is 24 minus integration from 0 to infinity similarly here x raised to 16 divided by 1 plus x raised to 24 can be written as x raised to 17 minus 1 here m is x raised to 16 can be written as x raised to 17 minus 1 that is here m is 17 and m plus n is 24 therefore n is 17 therefore we can write x raised to 17 minus 1 divided by 1 plus x raised to 17 plus 7 dx which is equal to we have beta of 7 17 minus beta of 17 7. By using the definition of beta function that is as a improper beta function as a improper integral which is equal to 0 because your beta of m n is equal to beta of n m. Now pause the video for a while and evaluate integration from 0 to infinity x raised to 8 divided by 1 plus x raised to 24 dx. I hope you have completed now see here we can write integration from 0 to infinity x raised to 8 divided by 1 plus x raised to 24 dx is equal to we can write integration from 0 to infinity x raised to 8 can be written as x raised to 9 minus 1 divided by 1 plus x raised to 9 plus 15 dx since here m is equal to 9 and m plus n is equal to 24 therefore what we get n is equal to 15. Now which is in the form of integration from 0 to infinity x raised to m minus 1 divided by 1 plus x raised to m plus n and which is equal to we can write beta of m n that is beta of 9 15. Now using the relation between beta and gamma function we can write which is equal to gamma of 9 into gamma of 15 divided by gamma of 9 plus 15. So gamma of 9 can be written as 8 factorial, gamma of 15 can be written as 14 factorial and gamma of 24 can be written as 23 factorial by using the property of gamma function this what are the value of the given integral. Now we will see one more example prove that integration from 0 to infinity root x divided by 4 plus 4x plus x squared dx is equal to pi by 2 root 2. Now consider the given example integration from 0 to infinity root x divided by 4 plus 4x plus x squared dx. Now which can be written as which is equal to integration from 0 to infinity x raised to 1 by 2 and this 4 plus 4x plus x squared can be written as 2 plus x whole square dx. And now we have to express this in terms of beta function as a improper integral and for that we can make use of the substitution x is equal to 2t therefore dx equal to 2dd. Accordingly here we have to change the limits when x is equal to 0 we get the value of t as a 0 and when x tends to infinity here t also tends to infinity. Therefore the given integral i we can write i equal to integration from 0 to infinity. Now here putting x equal to 2t that is 2 to raise to 1 by 2 divided by 2 plus 2t whole square and the value of dx as a 2dd which is equal to 2 raise to half into 2 means 2 root 2 and here you taking 2 common that is 2 square means what we get 4 into integration from 0 to infinity we get t raise to 1 by 2 divided by 1 plus t whole square dt. Now which is in the form of beta function. Therefore the given integral can be written as 1 by root 2 integration from 0 to infinity t raise to 1 by 2 can be written as t raise to 3 by 2 minus 1 divided by 1 plus t raise to 3 by 2 plus 1 by 2 dt since here we know m minus 1 is 1 by 2. Therefore the value of m is 3 by 2 and also here we have m plus n is 2 means here we have n is equal to 2 minus m that is 2 minus 3 by 2 which is equal to 1 by 2 that is here we have written the given integral as 1 by root 2 integration from 0 to infinity m minus 1 means 3 by 2 minus 1 divided by 1 plus t raise to m plus n that is 3 by 2 plus 1 by 2 means here m is 3 by 2 and n is 1 by 2. Therefore we can write which is equal to 1 by root 2 beta of m means 3 by 2 and n means 1 by 2. Now which is equal to here using the relation between beta and gamma function we can write beta of 3 by 2 and 1 by 2 as which is equal to 1 by root 2 gamma of 3 by 2 into gamma of 1 by 2 divided by gamma of 3 by 2 plus 1 by 2 which is equal to 1 by root 2 and gamma of 3 by 2 can be written as 1 by 2 into gamma of 1 by 2 keeping gamma of 1 by 2 as it is and gamma of 3 by 2 plus 1 by 2 is gamma of 2. Now which is equal to by simplifying this you get 1 by 2 root 2 and gamma of 1 by 2 whole square and gamma of 2 we can write 1 factorial which is equal to 1 by 2 root 2 and we know the gamma of value of gamma of 1 by 2 is root pi that is root pi whole square and gamma of 1 factorial is 1 and which is equal to we can write pi by 2 root 2 hence the root references higher engineering mathematics pi doctor B. S. Gravel.