 Hello. Myself. M.S. Basargaon. Assistant Professor. Department of Humanities and Sciences. Waltz and the Stopp Technologies. Ola. Go. In this video we are going to discuss about the gamma function part 1. Now, learning outcome. At the end of this session students will be able to evaluate definite integral by gamma function. Now, first of all we will see about the gamma function. What is gamma function? Gamma function is defined by the improper integral integration from 0 to infinity e raise to minus x into x raise to n minus 1 dx where n is greater than 0 and is denoted by gamma of n. Hence, gamma of n is equal to integration from 0 to infinity e raise to minus x into x raise to n minus 1 dx where n is greater than 0. It is a special kind of definite integral. Now, we will see the properties of gamma function. First important property is gamma of n plus 1 is equal to n into gamma of n. Now, we will prove this property by using the definition of gamma function. In the gamma function replacing here n by n plus 1 we get gamma of n plus 1 is equal to integration from 0 to infinity e raise to minus x into x raise to n dx. Integrating by parts we get here gamma of n plus 1 is equal to taking u as a x raise to n and v as a e raise to minus x keeping x raise to n as it is and integration of e raise to minus x is minus of e raise to minus x with a limit 0 to infinity minus integration from 0 to infinity derivative of x raise to n is n into x raise to n minus 1 and integration of e raise to minus x is minus of e raise to minus x dx. Since we know integration of u v dx is equal to u into integration of v dx minus integration of derivative of u with respect to x that is du by dx into integration of v dx dx. Therefore, gamma of n plus 1 is equal to no putting the upper limit here as x raise to infinity x raise to n by e raise to x raise to 0 therefore, upper limit is 0. Similarly, by putting x equal to 0 that is lower limit we get the value of this term will be 0. Therefore, the value of this term will be 0 plus now taking here minus minus that we get plus and taking n common that is n into integration from 0 to infinity e raise to minus x into x raise to n minus 1 dx. Therefore, we can write gamma of n plus 1 is equal to n into gamma of n here this what we get again we are getting the definition of gamma function. Therefore, gamma of n plus 1 is equal to n into gamma of n now this is known as recurrence or reduction formula of gamma function. Now, we will see take the note of again property of gamma of n plus 1 here gamma of n plus 1 is equal to n factorial if n is positive integer that is applying successfully this reduction formula we get gamma of n plus 1 is equal to n factorial if n is positive integer gamma of n plus 1 is equal to n into gamma of n if n is positive real number and gamma of n is equal to gamma of n plus 1 divided by n if n is negative fraction and one more important property of gamma function is gamma of 1 by 2 equal to root pi. Now, we will see the example find the value of gamma of minus 5 by 2 now here minus 5 by 2 is a negative fraction. Therefore, we can use the property gamma of n is equal to gamma of n plus 1 divided by n now here in this property putting n equal to minus 5 by 2 that is gamma of minus 5 by 2 is equal to gamma of minus 5 by 2 plus 1 divided by minus 5 by 2 which is equal to we can write minus 2 by 5 and gamma of minus 5 by 2 plus 1 is gamma of minus 3 by 2 which is equal to minus 2 by 5 again applying the same property we can write this as gamma of minus 3 by 2 can be written as gamma of minus 3 by 2 plus 1 divided by minus 3 by 2 which is equal to we get 4 by 15 into gamma of minus 1 by 2 which is equal to 4 by 15 I can again applying the same formula we get gamma of minus 1 by 2 plus 1 divided by minus 1 by 2 which is equal to we can write minus 8 by 15 gamma of 1 by 2 which is equal to minus 8 by 15 and again gamma of 1 by 2 is root pi. Now pause the video for a while and find the value of gamma of 6 and gamma of 5 by 2 I hope you have completed now see here gamma of 6 is equal to we can write 5 factorial that is 5 into 4 into 3 into 2 into 1 which is equal to 120 since we know gamma of n plus 1 is equal to n factorial and secondly gamma of 5 by 2 is equal to we can write gamma of 3 by 2 plus 1 since gamma of n plus 1 equal to we have n into gamma of n which is equal to 3 by 2 into gamma of 3 by 2 which is equal to we can write again 3 by 2 into gamma of 3 by 2 can be written as gamma of 1 by 2 plus 1 which is equal to 3 by 2 into 1 by 2 into gamma of 1 by 2 by simplifying this we get 3 by 4 and gamma of 1 by 2 is root pi. Now we will see the more example evaluate the integral integration from 0 to infinity e raise to minus root x into x raise to 1 by 4 dx. Now to express this function in terms of gamma function here we have to make use of the proper substitution that is here we can make use root x is equal to t that is x is equal to t square therefore, dx equal to 2t dt. Now here we are using the substitution for x as a function of t therefore, we have to find the limits of t now when x is equal to 0 putting x equal to 0 we get t equal to 0 and when x tends to infinity we get t tends to infinity. Therefore, integration from 0 to infinity e to the power minus root x into x raise to 1 by 4 dx is equal to integration from 0 to infinity here substituting root x equal to t that we get e raise to minus t and we are putting x equal to t square that is t square raise to 1 by 4 and dx as a 2t dt which is equal to taking 2 outside the integral that is 2 into integration from 0 to infinity that e raise to minus t and the value of t square raise to 1 by 4 into t is t raise to 3 by 2 dt. Now comparing this with the definition of gamma function here n minus 1 is 3 by 2 and which is equal to you can write twice gamma of 5 by 2 that is n is here 5 by 2 which is equal to 2 into 3 by 2 into 1 by 2 into gamma of half by using the property of gamma of n plus 1 equal to n into gamma of n which is equal to by simplifying it 3 by 2 and the value of gamma of half is root pi. Now we will see one more example evaluate the integral integration from 0 to infinity x square into e raise to minus h square x square dx. Now here again we have to make use of the proper substitution to express this in terms of gamma function and here we have to make use the substitution h square x square equal to t to get this integral in terms of gamma function that is we get x square is equal to t by h square that is x equal to t raise to half divided by h. Therefore dx is equal to 1 by h into 1 by 2 into t raise to minus half dt which is equal to 1 by 2 h t raise to minus half dt. Now again here we have to change the limits accordingly when x is equal to 0 we get t equal to 0 and when x raise to infinity t raise to infinity. Therefore integration from 0 to infinity x square into e raise to minus h square x square dx is equal to integration from 0 to infinity. Now here substituting the value of x square as t by h square into e raise to minus t and the value of dx is 1 by 2 h t raise to minus half dt. Now taking the constant term outside which is equal to 1 by 2 h cube into integration from 0 to infinity e raise to minus t and t raise to 1 by 2 dt. Now which is equal to 1 by 2 h cube here n minus 1 is 1 by 2 therefore n is 3 by 2 therefore we get gamma of 3 by 2 which is equal to 1 by 2 h cube gamma of 3 by 2 can be written as gamma of 1 by 2 plus 1 and by using again the property gamma of n plus 1 is equal to n into gamma of n we can write 1 by 2 h cube into 1 by 2 into gamma of 1 by 2 which is equal to 1 by 4 h cube and the value of gamma half half is root pi. Therefore the value of the integral is 1 by 4 h cube into root pi references higher engineering mathematics by Dr. B. S. Graywall.