2
$\begingroup$

Is the following statement true:

Let $g(x)$ be some non negative continuous function of $x$.We know that$$\int_{0}^\infty e^{-\beta x}x^{\alpha-1}dx={\Gamma(\alpha)\over \beta^\alpha}$$

Does the following hold as well: $$\int_{0}^\infty e^{-\beta\cdot g(x)}g(x)^{\alpha-1}dx={\Gamma(\alpha)\over \beta^\alpha}\ \ ?$$

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ The general statement is certainly false (consider $g\equiv 0$). Look at integration by substitution - assuming a piecewise differentiable $g$, you will need its derivative $g'(x)$. Dividing the integral into two pieces, $\int_{-\infty}^0$ and $\int_0^\infty$, it looks to me like the specific statement for $g(x)=|x|$ is also false. $\endgroup$
    – Stephan Kolassa
    Commented Oct 22, 2016 at 13:21
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because it has been wrongly posted: the claim made is trivial. $\endgroup$
    – Qwerty
    Commented Oct 22, 2016 at 13:34
  • $\begingroup$ @Xi'an Since it is answered with upvotes, I can't $\endgroup$
    – Qwerty
    Commented Oct 22, 2016 at 14:02
  • 1
    $\begingroup$ If you want a reasonably general result along those lines, to replace $x$ by $g(x)$ you need to replace $dx$ by $g'(x)\, dx$ (i.e. you need to keep the Jacobian of the transformation) --- and you need to also convert the limits on the integral when you transform. Then the substitution $u=g(x), du=g'(x) dx$ converts it back to an integral of the original form and you have a result similar to the one you describe. $\endgroup$
    – Glen_b
    Commented Oct 23, 2016 at 0:34

1 Answer 1

Reset to default
4
$\begingroup$

As stated, the relation is not true: consider $g(x) = 1$, in which case the integral is infinite.

Using the general framework of integration by substitution, the following conditions are sufficient:

  • $g$ is differentiable, with an integrable derivative
  • $g(0) = 0$ and $g(\infty) = \infty$
  • $g'(x) = 1$ for all $x$

This, unfortunately, only allows for the function $g(x) = x$.

Of course, this doesn't prove that the relation doesn't still hold for other $g$. But the most obvious way to show it doesn't work.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I edited my question. Please edit your answer so as to suit my new needs $\endgroup$
    – Qwerty
    Commented Oct 22, 2016 at 13:38
  • $\begingroup$ Nonnegative is still not enough; consider $g(x) = 1$. $\endgroup$
    – Danica
    Commented Oct 22, 2016 at 13:43
  • $\begingroup$ So, will it be possible to list down the exact conditions on $g$ such that this holds? $\endgroup$
    – Qwerty
    Commented Oct 22, 2016 at 13:53

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