 First, I myself, M.S. Basragaon, Assistant Professor, Department of Humanities and Sciences, Pulchandist of Technology, Solapur. In this session, we are going to discuss gamma function R2, learning outcome. At the end of this session, students will be able to apply gamma function to evaluate definite integrals. Pause the video for a while and evaluate integration from 0 to infinity e raised to minus x cube dx. I hope you have completed. So, in the first session, we have discussed about the gamma function. Now, here the given integral is integration from 0 to infinity e raised to minus x cube dx. Now, to express this in terms of gamma function, that is, integration from 0 to infinity e raised to minus x into x raised to n minus 1 dx, here the proper substitution is x cube is equal to t. Taking cube root on both sides, we get x equal to t raised to 1 by 3, that is, dx is equal to 1 by 3 t raised to minus 2 by 3 dt. Now, here, accordingly, you have to change the limit. When x is equal to 0, we get t equal to 0 and when x tends to infinity, t tends to infinity. Therefore, integration from 0 to infinity e raised to minus x cube dx is equal to integration from 0 to infinity e raised to minus 3 and the value of dx is 1 by 3 t raised to minus 2 by 3 dt, which is equal to 1 by 3 integration from 0 to infinity e raised to minus 3 into t raised to minus 2 by 3 dt, which is equal to we can write 1 by 3 gamma of 1 by 3, here comparing with the definition of gamma function n minus 1 is equal to minus 2 by 3. Therefore, n is equal to 1 minus 2 by 3, that is what, 1 by 3. Therefore, the value of the given integral is 1 by 3 into gamma of 1 by 3. Now, in this one, again, we will see the different types of example to be expressed in terms of gamma function. Now, here, take the example, evaluate integration from 0 to 1, x cube log of 1 by x raised to the power 4 dx. Now, here, the given integral involves logarithm function. Therefore, first of all, we will simplify here. Therefore, we can write integration from 0 to 1, x cube log of 1 by x raised to 4 dx is equal to integration from 0 to 1, x cube log of 1 by x raised to 4 can be written as 4 into log of 1 by x dx, since we know log of a raised to b as a b log a, which is equal to we can write 4 into integration from 0 to 1, x cube log of 1 by x dx. Now, we have to express this integral in terms of e raised to minus 3, for that the suitable substitution is let log of 1 by x is equal to t, that is, we can write 1 by x equal to e raised to t. Therefore, we can write x is equal to e raised to minus t. Therefore, dx is equal to minus e raised to minus t dt. Now, here, accordingly, we will change the limit when x is equal to 0. If we put x equal to 0, then in this one, then t tends to infinity and when x is equal to 1, that is, here log of 1 is 0, therefore, t equal to 0. Therefore, the given integral can be written as integration from 0 to 1, x cube log of 1 by x raised to 4 dx is equal to 4 into here, lower limit is infinity and upper limit is 0, that is, infinity to 0, e raised to minus 3 t into t and the value of dx is minus e raised to minus t dt, which is equal to we can write 4 into integration from 0 to infinity e raised to minus 4 t, that is, e raised to minus 3 t into e raised to minus t, that is, e raised to minus 4 t dt. Here, we are using the fundamental theorem of definite integral, that is, integration from a to b f of x dx is equal to minus integration from b to a f of x dx. If you change the upper and lower limit, change in the sign of the integral. Now, again, this function is not in terms of gamma function. Therefore, again, we have to make use of the substitution 4 t equal to y, that is, t equal to y by 4, therefore, dt equal to dy by 4. Again, here, we have to change the limit and t equal to 0, we get y equal to 0 and when t tends to infinity, y tends to infinity. Therefore, the given integral can be written as 4 into integration from 0 to infinity e raised to minus y, y by 4, dy by 4, which is here 4, 4 will get cancelled, taking 1 by 4 common outside, that is, 1 by 4 integration from 0 to infinity e raised to minus y into y dy, which is equal to 1 by 4 into gamma of 2, because here n minus 1 is 1. Therefore, n is 2. Therefore, we can write gamma of 2. Gamma of 2 can be written as 1 factorial by using n plus 1 equal to n factorial by 4, which is equal to 1 by 4. Now, we will see the next example. Evaluate the integral, integration from 0 to infinity e raised to minus b x square dx. Now, here base is a, we have to convert this to the base e and here to express this integrand in terms of exponential function, we have to make with the substitution e raised to minus b x square equal to e raised to minus t. Taking logarithm on both side, we get minus b x square log a equal to minus t log e and simplifying, we get b x square log of a equal to t, because log to the base e of e is 1. Now, solving for x square, here we get x square equal to t by b log a, that is, x equal to t raised to 1 by 2 divided by square root of b log a. Therefore, dx is equal to 1 by square root of b log a into 1 by 2 t raised to minus half dt. Now, accordingly, we have to change the limit when x is equal to 0, putting x equal to 0, here we get t equal to 0 and when x tends to infinity, t tends to infinity. Therefore, the given integral, integration from 0 to infinity e raised to minus b x square dx is equal to integration from 0 to infinity e raised to minus t and the value of dx as 1 by 2 root b log a into t raised to minus 1 by 2 dt. Now, taking constant term on both side, which is equal to 1 by 2 root b log a into integration from 0 to infinity e raised to minus t into t raised to minus half dt, which is equal to 1 by 2 root b log a gamma of 1 by 2, because here n minus 1 is minus 1 by 2. Therefore, n is 1 by 2, which is equal to root pi divided by 2 root b log a. This is what the value of the given integral. Now, we will see one more example of different type. Evaluate integration from 0 to infinity x raised to 5 by 5 raised to x dx. Now, here we have to convert this in terms of gamma function and for that, suitable substitution is here 5 raised to x is equal to e raised to t. Now, here taking logarithm on both side, we can write x log 5 is equal to t, because log of e is 1. Therefore, dx is equal to 1 by log 5 dt. Now, here again we have to change the limits, when x is equal to 0, putting x equal to 0, here we get the value of t equal to 0 and when x tends to infinity, t tends to infinity. Now, substituting the substitution and the limits in the given integral, that is integration from 0 to infinity x raised to 5 divided by 5 raised to x dx is equal to integration from 0 to infinity. Now, here we have to substitute the value of x, that is t by log 5 raised to the power 5 and 5 raised to x, that is e raised to t and the value of dx as 1 by log 5 dt. Now, taking the constant term, that is 1 by log 5 raised to 5 and 1 by log 5, that you can take outside the integral, which is equal to 1 by log 5 raised to 6 into integration from 0 to infinity e raised to taking e raised to t to the numerator, that we get e raised to minus t and t raised to 5 dt, which is equal to, we can write log 5 raised to 6 into gamma of 6, because here n minus 1 is 5, therefore, n is 6. Therefore, the value of this, we can write gamma of 6 and which is equal to, we can write the value of gamma of 6, it can be written as 5 factorial by using the property gamma of n plus 1 is equal to n factorial if n is positive integer. Therefore, we can write here the value of gamma of 6 as a 5 factorial divided by log 5 raised to 6. Here 5 factorial means 120, which is equal to 120 divided by log 5 raised to 6, references higher engineering methods by Dr. B. S. Gravel.