Proving the Gamma Function: Using Substitution and Integral Calculus

[iceman]

Hello, can anyone please me here?

I need to prove that

int(x^a(lnx)^b.dx= (-1)^b/((1+a)^b+1)*Gamma(b+1)

by making the substitution x=e^-y

this is what I have done so far:

x=e^-y -> y=-lnx

x=0 -> y=-(-00) =+00
x=1 -> y=0

dy/dx = -1/x -> dx=-xdy =-e^-ydy

then the integral becomes

int[e^(-ay)*(-y)^b*(-e^-y)dy, lower lim->+00, upper lim-> 0
= (-1)^b*int[e^-(a+1)y*y^bdy.

I then made a substituion t=(a+1)y
so integral becomes

(-1)^b*int[e^-t*y^bdy]

this is where I get a little bit lost...!

 

[Hurkyl]

Try the substitution with t again.

 

[M98Ranger]

To continue, we can use integration by parts. Let u = y^b and dv = e^-t dt. Then du = b*y^(b-1) and v = -e^-t. The integral becomes:

(-1)^b*int[u*dv]
= (-1)^b*[u*v - int[v*du]] (integration by parts formula)
= (-1)^b*[y^b*(-e^-t) + b*int[e^-t*y^(b-1)]]
= (-1)^b*[y^b*(-e^-t) + b*u*(a+1)^(b-1)] (using the substitution t=(a+1)y)
= (-1)^b*[y^b*(-e^-t) + b*y^b*(a+1)^(b-1)]

Now, we can substitute back in our original variables to get:

(-1)^b*int[e^-t*y^bdy]
= (-1)^b*[y^b*(-e^-t) + b*y^b*(a+1)^(b-1)]
= (-1)^b*[x^b*e^-(-lnx) + b*x^b*(a+1)^(b-1)]
= (-1)^b*[x^b*x + b*x^b*(a+1)^(b-1)]
= (-1)^b*x^(b+1) + b*x^b*(a+1)^(b-1)

Finally, we can substitute in our limits of integration (0 and +00) to get:

(-1)^b*x^(b+1) + b*x^b*(a+1)^(b-1), lower lim->+00, upper lim-> 0
= (-1)^b*[+00^(b+1) + b*0^b*(a+1)^(b-1)]
= 0 - 0 = 0

Therefore, our final result is:

int(x^a(lnx)^b.dx= (-1)^b/((1+a)^b+1)*Gamma(b+1)

which proves the original statement. I hope this helps!