In Spivak's Calculus 19 – 44, I'm being asked to prove that the previously encountered gamma function, defined as: $$\Gamma(x)=\int_0^\infty e^{-t}t^{x-1}\,dt$$
is equivalent to
$$\frac{1}{x}\int_0^\infty e^{-u^{1/x}}\,du$$
with the use of the substitution
$u=t^x$.
However I can't seem to get the lower limit of integration, $0$.
Here's my argument:
$$\int_0^\infty e^{-t}t^{x-1} \, dt=\frac{1}{x}\int_0^\infty e^{-t}xt^{x-1} \, dt$$
let $g(x)=t^x$, $g'(x)=xt^{x-1}$, $f(u)=e^{u^{1/x}}$ and we get:
$$\frac{1}{x}\int_{g(0)}^{g(\infty)}e^{-u^{1/x}} \, du=\frac{1}{x} \int_1^\infty e^{-u^{1/x}} \, du$$
with the use of the substitution formula.
What am I missing? how do I get the 0?
Here's how the author does it:
Let $u=t^x$, $du=xt^{x-1} \, dt$. Then
\begin{align}
\Gamma(x)&=\int_0^\infty e^{-t}t^{x-1} \, dt=\int_0^\infty e^{-u^{1/x}}\frac{du}{x} \\
&=\frac{1}{x}\int_0^\infty e^{-u^{1/x}} \, du
\end{align}
But I don't understand how he can get away with not substituting the g function into the limits of integration, i feel stupid.
Thank you all in advance, even if I can't participate in this forum yet I want you to know that you've helped me a lot.
Simone
Edit: SOLVED! the argument of the $g$ function should be $t,$ not $x$!
Mind your "arguments", guys.
Best Answer
I have found the point of error. The argument of the $g$ function should be $t$, not $x$.