How do I prove that $\displaystyle \int_{E^n}|\mathbf{x}|^{-|\mathbf{x}|}d\lambda(\mathbf{x})$, where $\lambda$ is lebesgue measure, converges?
I was thinking of finding an upper bound function for $|\mathbf{x}|^{-|\mathbf{x}|}$ and then use the comparison test.
However I don't seem to be able to find that function.
Any help would be appreciated. Thanks.
Edit: As an addendum to a response, I cannot change variables. I'm supposed to do it without that technique.
By $E^n$ you mean $\mathbb R^n$? If yes, you can use polar coordinates, and write $$\int_{\mathbb R^n}|x|^{-|x|}\,dx=\int_0^{\infty}r^{n-1}\int_{S^{n-1}}r^{-r}\,dx'\,dr=|S^{n-1}|\int_0^{\infty}r^{n-1-r}.$$ Break the integral from $0$ to $n+1$, and there you get a continuous function, so it converges, while on $(n+1,\infty)$: $$n-1-r\leq n-1-(n+1)=-2\Rightarrow \int_{n+1}^{\infty}r^{n-1-r}\,dr\leq\int_{n+1}^{\infty}r^{-2}\,dr<\infty.$$ So, the original integral converges.
Edit: Without changing coordinates, you can do the following: let $B_m$ be the ball centered at $0$, with radius $m$. Then, write $$\int_{\mathbb R^n}|x|^{-|x|}\,dx=\int_{B_1}|x|^{-|x|}\,dx+\sum_{m=1}^{\infty}\int_{B_{m+1}\smallsetminus B_m}|x|^{-|x|}\,dx.$$ Since $f(x)=|x|^{-|x|}$ is continuous on $B_1$ (if you set $f(0)=1$), the integral of $f$ over $B_1$ converges. Now, if $m\geq 1$: if $x\in B_{m+1}\smallsetminus B_m$, you have that $m\leq|x|\leq m+1$, therefore $$\int_{B_{m+1}\smallsetminus B_m}|x|^{-|x|}\,dx\leq\int_{B_{m+1}\smallsetminus B_m}(m+1)^{-m}\,dx\leq|B_{m+1}|(m+1)^{-m}.$$ But, the series $\displaystyle \sum_{m=1}^{\infty}|B_{m+1}|(m+1)^{-m}$ converges: you have that $$\frac{|B_{m+2}|(m+2)^{-m-1}}{|B_{m+1}|(m+1)^{-m}}=\frac{|B_1|(m+2)^n(m+2)^{-m-1}}{|B_1|(m+1)^n(m+1)^{-m}}=\left(1+\frac{1}{m+1}\right)^n\left(1+\frac{1}{m+1}\right)^{-m}\frac{1}{m+2},$$ which converges to $1\cdot e^{-1}\cdot 0=0$. So, from the ratio test, this series converges. Therefore, the integral of $f$ over $\mathbb R^n$ converges.