$f(x_1,x_2)=\int_{0}^{\sqrt{{x_1}^2+{x_2}^2}}e^{-\frac{w^2}{{x_1}^2+{x_2}^2}}dw$ is homogeneous of which degree? Problem: Consider the function $$f(x_1,x_2)=\int_{0}^{\sqrt{{x_1}^2+{x_2}^2}}e^{-\frac{w^2}{{x_1}^2+{x_2}^2}}dw$$ with the property that $f(0,0)=0$. Then the function $f(x_1,x_2)$ is homogeneous of degree -1 or homogeneous of degree 1/2 or homogeneous of degree 1?
My attempt: I took the substitution $$t=\frac{w^2}{x_1^2+x_2^2}$$ and the integral turned out to be $$f(x_1,x_2)=\frac{(x_1^2+x_2^2)^{3/2}}{2}\int_{0}^{1}\frac{e^{-t}}{\sqrt{t}}dt.$$ I am stuck here as I cannot compute the integral. I guess there is a way without computing the integral.
P.S.: There is another question in this site that is exactly the same as this one, but that page does not have any answer.
 A: Note that 
$$ \int_{0}^{\infty} \left( \frac{e^{-t}}{\sqrt{t}} \right) dt =  $$ 
$$ \int_{0}^{\infty} \left( e^{-t}t^{-\frac{1}{2}} \right) dt = \Gamma\left(\frac{1}{2}\right)$$ 
But your problem has a variable bound which can be handled by the 
https://en.wikipedia.org/wiki/Incomplete_gamma_function
So the integral becomes
$$ f(x_1,x_2) = \frac{(x_1^2 + x_2^2)^\frac{3}{2}}{2} \gamma\left(\sqrt{x_1^2+x_2^2},\frac{1}{2}\right)$$ 
Now it naturally follows that 
$$ f(cx_1,cx_2) = c^3 \left(\frac{(x_1^2 + x_2^2)^\frac{3}{2}}{2} \gamma\left(c\sqrt{x_1^2+x_2^2},\frac{1}{2}\right)\right)$$ 
I would presume something slightly greater than degree 3 would be in order. Or even non linear degree (but slightly larger than 3). Not sure how to get the c outside of the gamma though.
After editing the bound to 0 to 1 the integral must be CONSTANT. So it follows that the function is homogenous with degree 3.
A: Hoping that I am not off topic, you face the problem of $$I=\int e^{-a w^2}\,dw=\frac{\sqrt{\pi } }{2 \sqrt{a}}\text{erf}\left(\sqrt{a} w\right)$$ where appears the error function. Then $$J=\int_0^b e^{-a w^2}\,dw=\frac{\sqrt{\pi } }{2 \sqrt{a}}\text{erf}\left(\sqrt{a} b\right)$$ So, now $a=\frac 1{x_1^2+x_2^2}$, $b=\sqrt{x_1^2+x_2^2}$ then lead to $$f(x_1,x_2)=\int_{0}^{\sqrt{{x_1}^2+{x_2}^2}}e^{-\frac{w^2}{{x_1}^2+{x_2}^2}}\,dw=\frac{ \sqrt{\pi }}{2} \sqrt{x_1^2+x_2^2}\,\,\text{erf}(1) $$ $$f(cx_1,cx_2)=c\,f(x_1,x_2)$$
