Integration (fromula of reduction)

[Miike012]

Referring to the paint document. Could they have easily changed m into m + 2 or m + 1.5? Or does it have to contain the variable m and n?

 

[Miike012]

Also, in the original equation (Equation 1 of the paint document) why did they chose the power of x to be (m-1)? Why not just leave it as m?

Therefore we would get the equation

∫x^mX^pdx = X^px^m+1/(m+1) - bnp/(m+1)∫x^{m+n + 1}X^p-1dx

and therefore if we wanted to solve for the integral of say x²√(1+x) then the value of m is 2 and the value of n is 1 and the value of p is 1/2

which saves you a step in solving for m: m-1 = 2 and m = 3, if we used the equation that they started with which was

∫x^mX^pdx = X^px^m+1/(m+1) - bnp/(m+1)∫x^{m+n + 1}X^p-1dx where m is equal to m-1

 

[haruspex]

Referring to the paint document. Could they have easily changed m into m + 2 or m + 1.5? Or does it have to contain the variable m and n?

Not sure I understand your question.
You can make any of those substitutions as long as you are consistent. They chose m-n because it gave the form they were after on the LHS.
Wrt m v. m-1, using m-1 on the LHS of (1) makes the equation as a whole slightly simpler.

 

[Miike012]

Not sure I understand your question.
You can make any of those substitutions as long as you are consistent. They chose m-n because it gave the form they were after on the LHS.
Wrt m v. m-1, using m-1 on the LHS of (1) makes the equation as a whole slightly simpler.

Say I was asked to solve the integral of x²/(x^2 + 1)^1/2

then I have:

∫x²/(x^2 + 1)^1/2dx = ∫d{(x²+1)}^1/2/dx*x*dx
=(x²+1)^1/2*x - ∫(x²+1)dx.

On the LHS can I change the power of x in the numerator from 2 to 2 - m so long as I make the correct changes on RHS?

Because that is what they essentially did

 

[haruspex]

On the LHS can I change the power of x in the numerator from 2 to 2 - m so long as I make the correct changes on RHS?

Because that is what they essentially did

No, changing a constant to a variable is rather more drastic. You might have made use of specific attributes of the constant in arriving at the equation. All they did was a change of variable, but using the same name for the new variable.

 

[Ray Vickson]

Say I was asked to solve the integral of x²/(x^2 + 1)^1/2

then I have:

∫x²/(x^2 + 1)^1/2dx = ∫d{(x²+1)}^1/2/dx*x*dx
=(x²+1)^1/2*x - ∫(x²+1)dx.

On the LHS can I change the power of x in the numerator from 2 to 2 - m so long as I make the correct changes on RHS?

Because that is what they essentially did

No, it is not what they did---and you cannot do it, either. 2 is 2 and you cannot re-write it as something else.

In expanded form, what they did was, essentially, to first re-name m as m'-n, and rename p as p'+1, then remove the primes so that they can write m instead of m' and p instead of p'. They are not actually changing anything; they are simply re-naming things.