# I Miscellaneous Integration Technique

1. Mar 20, 2016

### Noriele Cruz

We were taught of several integration technique, only to find one of those techniques came up as years of solving of our professor.. Can someone explain to me how substitution

x = 1 / z
dx = -dz / z^2

works for some problems?
He called this reciprocal substitution, as what you can literally see.

2. Mar 20, 2016

### pwsnafu

Consider the inverse gamma distribution, with pdf
$f(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{-\alpha-1} e^{-\beta / x}$.
If you want to integrate this over say [0,a], doing the transform converts the integral into the standard gamma distribution, which is easier to work with.

3. Apr 9, 2016

### BreCheese

I don't see a substitution, or perhaps I don't understand your question

What your professor did is apply the power rule, which states that if y=x^n, then y'=nx^(n-1)
So in your particular problem this is the step-by-step way to get the derivative:
x=1/z
x=z^(-1)
dx/dz=(-1)z^(-1-1) ; read, the derivative of x, with respect to z equals negative one z to the power of the quantity negative one minus one.
dx/dz=(-1)z^(-2)
dx/dz= -1/z^(2) ; now multiple each side of the equation by "dz" to get rid of the dz on the left side of the equation
dz(dx/dz)= (-1/z^(2))dz
dx= -dz/z^(2)

I wish I knew how to use LaTex so I could be more clear.
In short, the notation for writing a derivative is the following
y'(x), read "y-prime of x" is the same as saying:
dy/dx , read "the derivative of y with respect to x"

When we make a u substitution during integration we have to remember to replace our original "with respect to..." statement with the new "with respect to u" statement.
Example:
∫(x/(x^(2)+1))dx
u=x^(2)+1
du/dx=2x read, "the derivative of u with respect to x
du=(2x)dx
dx=du/2x Now substitute this back into the original integral:
∫(x/(x^(2)+1))dx now equals ∫(x/u2x)du
the x in the numerator cancels with the x in the denominator so the integral is now:
∫(1/2u)du Now bring the 1/2 out front of the integral:
1/2∫(1/u)du Now integrate:
(1/2)ln(u)+C Now substitute "u" back in
(1/2)ln(x^(2)+1)+C