# IS $\int_{-\infty}^\infty e^{-\beta\cdot g(x)}g(x)^{\alpha-1}\text{d}x={\Gamma(\alpha)\over \beta^\alpha}\ \ ?$ [closed]

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

## closed as off-topic by Qwerty, Xi'an, mdewey, gung - Reinstate Monica♦, whuber♦Oct 22 '16 at 23:11

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• 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. – S. Kolassa - Reinstate Monica Oct 22 '16 at 13:21
• I'm voting to close this question as off-topic because it has been wrongly posted: the claim made is trivial. – Qwerty Oct 22 '16 at 13:34
• @Xi'an Since it is answered with upvotes, I can't – Qwerty Oct 22 '16 at 14:02
• 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. – Glen_b Oct 23 '16 at 0:34

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.

• I edited my question. Please edit your answer so as to suit my new needs – Qwerty Oct 22 '16 at 13:38
• Nonnegative is still not enough; consider $g(x) = 1$. – Dougal Oct 22 '16 at 13:43
• So, will it be possible to list down the exact conditions on $g$ such that this holds? – Qwerty Oct 22 '16 at 13:53