Find a function $\phi:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ which fixes the area under $f(x)=x^a e^{-\phi(a)x}$ for all $a>0$.
My investigation so far has lead me to believe that no such function exists, although this seemed unlikely to me at first.
I suppose we need to assume that $\phi$ can give a well defined area, and such $\phi$ certainly exist (e.g: $\phi(a)=1$).
Let $A(a)=\displaystyle\int_0^\infty x^a e^{-\phi(a)x}dx$ be the area under $f$. The substitution $u=\phi(a)x$ gives $A(a)=\frac{\Gamma(a+1)}{\phi(a)^{a+1}}$ where $\Gamma(z)=\displaystyle\int_0^\infty x^{z-1}e^{-x}dx$ is the gamma function. Then setting $A'(a)=0$ we obtain: $$\frac{d}{da}\log\phi(a)=\frac{\phi'(a)}{\phi(a)}=\frac{1}{a+1}\cdot\frac{\Gamma'(a+1)}{\Gamma(a+1)}=\frac{\psi(a+1)}{a+1}$$ where $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ is the digamma function, which has the series expansion (for $x>-1$): $$\psi(x+1)=-\gamma+\sum_{n=1}^{\infty}\frac{x}{n(x+n)}$$ where $\gamma$ is the Euler-Mascheroni constant. So, after a few partial fraction calculations we obtain: $$\begin{aligned} \log\phi(a)&=\int\left[\frac{1-\gamma}{a+1}-\frac1{(a+1)^2}+\sum_{n=2}^\infty\frac1{n(n-1)}\left(\frac{n}{a+n}-\frac{1}{a+1}\right)\right]da\\ &=(1-\gamma)\log(a+1)+\frac1{a+1}+\sum_{n=2}^\infty\left(\frac{\log(a+n)}{(n-1)}-\frac{\log(a+1)}{n(n-1)}\right)\\ \implies\phi(a)&=(a+1)^{1-\gamma} e^{\frac1{a+1}}\cdot \prod_{n=2}^\infty \frac{(a+n)^{\frac1{n-1}}}{(a+1)^{\frac1{n(n-1)}}}\\ &=(a+1)^{1-\gamma-\delta}e^{\frac1{a+1}}\cdot\prod_{n=2}^\infty (a+n)^{\frac1{n-1}} \text{ where }\delta=\sum_{n=2}^\infty\frac1{n(n-1)}<\infty\end{aligned} $$
However $\displaystyle\prod_{n=2}^\infty (a+n)^{\frac1{n-1}}>\prod_{n=2}^\infty (a+1)^{\frac1{n-1}}=(a+1)^{\sum_{n=1}^\infty\frac1n}=\infty$ since $(a+1)>1$ for $a>0$ and $\displaystyle\sum_{n=1}^\infty\frac1n = \infty$, so $\phi(a)$ is divergent for any $a>0$.
I suppose my question is thus: is my reasoning here correct, can no such function $\phi$ exist? And if not, what does such a function look like? Altnernatively, is there a simpler way to show what I've shown?