Let $\displaystyle f(x) = \beta\cdot \frac{1}{x^{\beta+1}}$ with $\beta > 0$
Asked for is the following integral: $\displaystyle \int_1^\infty f(x) \quad dx$
Now I wanted to use integration by substitution with the inverse of $f$, $f^{-1}(x)$.
(By defining $f: (0,\infty) \rightarrow (0,\infty)$ the functions should match the prerequisites.)
Here's the equality for integration by substitution (from Wiki):
$\displaystyle \int_{a}^{b} f(\varphi(t)) \cdot \varphi'(t)\,\mathrm{d}t = \int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x $
I now set $\,\,\varphi = f^{-1}$.
$\displaystyle \int_{a}^{b} f(f^{-1}(t)) \cdot \big(\frac{\mathrm{d}}{\mathrm{d}t} f^{-1}(t)\big)\,\mathrm{d}t = \int_{f^{-1}(a)}^{f^{-1}(b)} f(x)\,\mathrm{d}x $
$\Leftrightarrow$
$\displaystyle \int_{f(a)}^{f(b)}t \cdot \big(\frac{\mathrm{d}}{\mathrm{d}t} f^{-1}(t)\big)\,\mathrm{d}t = \int_{a}^b f(x)\,\mathrm{d}x $
However, when I now try to use the result on my problem, $f(\infty)$ tends to $0$, making it an integral $\displaystyle \int_1^0...$, which results in $0$, but should result in $1$.
We have $f^{-1}(t)=\beta^{1/(\beta+1)}t^{-1/(\beta+1)}$, so $\dfrac{d}{dt}f^{-1}(t)=-\beta^{1/(\beta+1)}\dfrac{1}{\beta+1}t^{-1/(\beta+1)-1}$, there is a negative sign here, so \begin{align*} \int_{f(a)}^{f(b)}t\cdot\dfrac{d}{dt}f^{-1}(t)dt&=\int_{\beta}^{0}t\cdot-\beta^{1/(\beta+1)}\dfrac{1}{\beta+1}t^{-1/(\beta+1)-1}dt\\ &=\int_{0}^{\beta}t\cdot\beta^{1/(\beta+1)}\dfrac{1}{\beta+1}t^{-1/(\beta+1)-1}dt\\ &=\dfrac{\beta^{1/(\beta+1)}}{\beta+1}\int_{0}^{\beta}t^{-1/(\beta+1)}dt\\ &=\dfrac{\beta^{1/(\beta+1)}}{\beta+1}\dfrac{1}{1-1/(\beta+1)}t^{1-1/(\beta+1)}\bigg|_{t=0}^{t=\beta}\\ &=\dfrac{\beta^{1/(\beta+1)}}{\beta+1}\dfrac{\beta^{1-1/(\beta+1)}}{1-1/(\beta+1)}\\ &=\dfrac{\beta}{\beta+1-1}\\ &=1. \end{align*}