I need to find the maximum of the Mellin transform $K(s)$ of the function $k(x)$, whose definition is $$K(s)=\int_0^\infty x^{s-1}k(x)dx$$ The answer is that the maximum of $K(s)$ is at $s=1$. But I have problems to derive it.
My attempt: In order to find the maximum I need to find the $s$ that satisfy $$K'(s)=0.$$ So I need to take the derivative of the integral above. At first, I thought I should apply the 1st Fundamental Theorem of Calculus, but I'm not so sure if this theorem is useful here since this a definite integral. By this theorem, if $$F(t)=\int_a^t x^{s-1}k(x)dx$$ then $$F'(t)=t^{s-1}k(t)$$ Roughly speaking, we derive wrt $x$ and then we substitute $x$ by $t$ to obtain $F'(t)$: $$\int \frac{d}{dx}x^{s-1}k(x)dx=\int d(x^{s-1}k(x))=x^{s-1}k(x) \to F'(t)=t^{s-1}k(t) $$ But it is not the same case as $K(s)$, because in this integral we have sumed over all the $x's$, then the result is a function that depends only on $s$, and we have to derive $K(s)$ wrt $s$, whereas in $F(t)$ the derivative is wrt $x$ . That is why I think this theorem doesn't work out here.
I also tried to prove it by the chain rule: $$\frac{d}{ds}K(s)=\frac{d}{ds}\int_0^\infty x^{s-1}k(x)dx=\int_0^\infty \frac{d}{dx} ( x^{s-1}k(x)) \frac{dx}{ds} dx =\int_0^\infty [ (s-1)x^{s-2}k(x)+x^{s-1}k'(x)]\frac{dx}{ds} dx$$ But then I don't see how this integrand would be equal to zero. If the second term of the integrand wasn't there, we'd obtain the answer. So I am wondering whether, as $s$ only appears in $x^{s-1}$, the chain rule only affect this term, I mean:
$$\frac{d}{ds}K(s)=\frac{d}{ds}\int_0^\infty x^{s-1}k(x)dx=\int_0^\infty \frac{d}{dx} ( x^{s-1} \frac{dx}{ds}) k(x) dx = \int_0^\infty (s-1)x^{s-2}\frac{dx}{ds} k(x) dx $$ The problem is that, as the derivative wrt $s$ doesn't affect $k(x)$, I am assuming that $x$ doesn't depend on $s$, then $dx/ds=0$, and $K'(s)$ would be zero for all $s$.
I would appreciate if you could point out the serious mistakes I've commited as much as any help you could bring me to solve this problem.