Undergrad Miscellaneous Integration Technique

[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.

 

[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.

 

[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

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