Integration of the exponential radial function Can anyone please help me regarding the following integral estimate.
Suppose  $B(0,R)=\{x\in\mathbb{R}^N:|x|<R\}$ and $\alpha>0$ is some positive constant. Then does the integral 
$$
I=\int_{B(0,R)}\,e^{\alpha|x|}\,dx
$$
satisfies the estimate:
$I\leq C\,R^{\beta}$ for some constants $\beta$, $C$ (both independent of $R$). Can you explicitly calculate $\beta$ if that estimate holds.
Thank you very much in advance... 
 A: No. If you use spherical coordinates then $dx=r^{N-1}drd\Omega$ where $d\Omega$ is the solid angle. Then you can write $I=\Omega\int_0^R e^{\alpha r}r^{N-1}dr $ and you can bound this as $I<\Omega e^{\alpha R}\frac{R^{N}}{N}$.
Notice also that if you expand the last integrand and integrate term by term you get $$I=\Omega \sum_{m=0}^\infty \frac{\alpha^m R^{m+N}}{m!(m+N)}$$ and this clearly is not bounded by $R^\beta$ since the sum is infinite.
A: I dont think it is even possible to get such an estimate. Rewriting 
$$
I=\int_0^R \int_{\partial B_r(0)}e^{\alpha |x|} dS dr = C(n)\int_0^R r^{n-1}e^{\alpha r}dr$$
where $C(n)$ is a constant associated with the dimension and the measure of the unit sphere. 
For your estimate to hold, we require that:
$$
C(n)\int_0^R r^{n-1}e^{\alpha r}dr \leq C R^{\beta}
$$
We can write the integrand as:
$$
f(r)=\sum_0^{\infty} \frac{\alpha^k r^{k+n-1}}{k!}
$$
And integrating, provided everything converges, gives 
$$
F(r)=\sum_0^{\infty} \frac{\alpha^k r^{n+k}}{k!(n+k)}
$$
Plugging in the values gives:
$$
\sum_0^{\infty} \frac{\alpha^k R^{n+k}}{k!(n+k)}-F(0)
$$
Assuming the desired estimate would hold, this would imply that
$$
\sum_0^{\infty} \frac{\alpha^k R^{n+k}}{k!(n+k)}-F(0) \leq C_2(n) R^{\beta}
$$ 
Sending $R \to \infty$ however, makes this claim false since we can always find $n+k \geq \beta$ and therefore $R^{n+k} \geq R^{\beta}$
