Integral of a scaled function Let $f$ be a positive function that integrates to 1 (i.e. a probability density function), that is:
\begin{align*}
\int_{-\infty}^{\infty}f(x)dx = 1.
\end{align*}
My intuition tells me that for any $\alpha>0$,
\begin{align*}
\int_{-\infty}^{\infty}f(\alpha x)dx = \int_{-\infty}^{\infty}f(x)dx =1,
\end{align*}
because as $x$ goes from $-\infty$ to $\infty$, then $\alpha x$ goes from $-\infty$ to $\infty$, and each $x$ in the integral has the same ''weight'' $dx$. 
However, by substitution it is easy to show that
\begin{align*}
\int_{-\infty}^{\infty}f(\alpha x)dx = \frac{1}{\alpha}.
\end{align*}
Can anyone explain this to me in words? 
 A: Perhaps this is easier to see with $$f(x) = \begin{cases} 1 &\text{if } 0 \leq x \leq 1 \\ 0 &\text{otherwise} \end{cases}$$ 
Clearly, the graph of $f$ is just one square with height and width $1$, so $\int_{-\infty}^{\infty} f(x) = 1$. 
However, $$f(\alpha x) = \begin{cases} 1 &\text{if } 0 \leq \alpha x \leq 1 \\ 0 &\text{otherwise} \end{cases}$$
Now, suppose that $\alpha > 0$. Then we can rewrite this as
$$f(\alpha x) = \begin{cases} 1 &\text{if } 0 \leq x \leq \frac{1}{\alpha} \\ 0 &\text{otherwise} \end{cases}$$
Now the graph of $f(\alpha x)$ is just one rectangle with height $1$ and width $\frac{1}{\alpha}$, so $\int_{-\infty}^{\infty} f(\alpha x) = \frac{1}{\alpha}$. 
The general case for other functions goes quite similiar, excepted that you do not have one rectangle that goes from width $1$ to width $\frac{1}{\alpha}$, but (possibly) infinitely many. 

Also, please note that the result at the end of your question is wrong when $\alpha<0$. It should be: \begin{align*}
\int_{-\infty}^{\infty}f(\alpha x)dx = \frac{1}{|\alpha|}.
\end{align*}
And I think one would commonly show this with substitution, not by integration by parts. 
A: Say $f(x) = \dfrac{1}{\pi(1+x^2)}$. So if we integrate it from $-\infty$ to $+\infty$, we will have:
$\int_{-\infty}^{+\infty}{\dfrac{1}{\pi(1+x^2)}}=\lim_{b\to+\infty}{\dfrac{\arctan(b)-\arctan(-b)}{\pi}}=\dfrac{\pi/2-(-\pi/2)}{\pi}=\dfrac{\pi}{\pi}=1$
So what happens when we integrate $f(2x)$? Well doing this will compress the function horizontally, by a factor of 2. So the shape of the initial graph(the red one) will change into the blue graph.(f(2x)) Thus the integration result will be $\dfrac{1}{2}$

So by multiplying $x$ by $\alpha>0$ in $f(x)$, you're actually compressing the function(or stretching it if $\alpha<1$) which will definitely change the area under the curve.(Unless $\alpha = 1$)
