1
$\begingroup$

I am self teaching myself python and computational physics via Mark Newmans book Computational Physics the exercise is 5.17 of Computational Physics. I have to shift the limits of integration for an improper integral after a change of variables so that Gaussian quadrature method gives a more accurate result. I am suppose to approximate the Gamma function $\Gamma(a)$

The relevant equation is here: enter image description here

This is the question:

enter image description here enter image description here

When I isolate for $x$ I get $x = \frac{zc}{1-x}$. But when $z = \frac{1}{2}$ I get $x = c$ but this makes no sense because this implies z is a constant which it's not. I know in order to shift a function $f(x)$ to the left you $f(x + c)$ but how would I do this with an integrand?

$\endgroup$
1
$\begingroup$

The task in this problem is to apply the variable transformation $$z=\frac{x}{c+x} \Leftrightarrow x=\frac{zc}{1-z}\,,\quad \textrm{d}x=\frac{c}{(1-z)^2}\textrm{d}z$$ to the integral $$\int\limits_0^\infty f(x)\textrm{d}x\,,\quad f(x)=x^{a-1}e^{-x}\,,$$ which yields $$\int\limits_0^1 \tilde{f}(z)\textrm{d}z\,,\quad \tilde{f}(z)=\left(\frac{zc}{1-z}\right)^{a-1}\exp\left(-\frac{zc}{1-z}\right)\frac{c}{(1-z)^2}\,,$$ and chose the parameter $c$ such that $\tilde{f}(z)$ takes its maximum at $z=\frac{1}{2}$.

How do we do that?

You have already completed the first part of the question: $z$ takes the value $\frac{1}{2}$ when $x=c$.

The second part asks which is the correct $c$. We tackle this by computing the derivative of $\tilde{f}(z=\frac{1}{2})$ with respect to the unknown parameter $c$ and set this to zero: $$\left.\frac{\partial \tilde{f}}{\partial c}\right\vert_{z=\frac{1}{2}} = \frac{1}{4}c^{a-1}e^{-c}\left(a-c\right)=0\,.$$ Thus, we find $c=a$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.