If the Laplace transform of the pdf of a random variable $X$ is given as $$\exp(Cs^{\frac{2}{m}})$$ then how can we show that the complementary cumulative distribution function of $X$ is proportional to $C$ where $C$ is some real value (actually $C=\text{some positive real value}\times \Gamma \left(-\frac{2}{m}\right)$) and $m>2$. Any intuitive or mathematical answer will be very helpful. Thanks in advance.
My Try:
If the Laplace transform of the pdf of a random variable ($X$) is $L(s)$ then the Laplace transform associated with the CCDF of that random variable is $L'(s)=\frac{1}{s}-\frac{L(s)}{s}$ and the corresponding CCDF of $X$ is given by Euler characterization as follows $$P(X>y)=\frac{2e^{ay}}{\pi}\int_0^{\infty}Re \left(L'(a+iu)\right)\cos(uy)du$$ For simplification I am going to use $a=0$ then the above formula becomes $$P(X>y)=\frac{2}{\pi}\int_0^{\infty}Re \left(L'(iu)\right)\cos(uy)du$$ using $L'(iu)=\frac{1}{iu}-\frac{\exp(C(iu)^{\frac{2}{m}})}{iu}$ asnd using the fact that $i=\exp(i\frac{\pi}{2}), \exp(ix)=\cos(x)+i\sin(x)$ we can simplify $L'(iu)$ as follows $$L'(iu)=\frac{1}{iu}+\frac{i}{u}\exp\left(Cu^{\frac{2}{m}}\cos(\frac{\pi}{m})\right)\cos\left(Cu^{\frac{2}{m}}\sin(\frac{\pi}{m})\right)-\frac{1}{u}\exp\left(Cu^{\frac{2}{m}}\cos(\frac{\pi}{m})\right)\sin\left(Cu^{\frac{2}{m}}\sin(\frac{\pi}{m})\right)$$ and correspondingly the real part will be $$Re\left(L'(iu)\right)=-\frac{1}{u}\exp\left(Cu^{\frac{2}{m}}\cos(\frac{\pi}{m})\right)\sin\left(Cu^{\frac{2}{m}}\sin(\frac{\pi}{m})\right)$$Putting this value in the formula of $P(X>y)$ we can write $$P(X>y)=-\frac{2}{\pi}\int_0^{\infty}\frac{1}{u}\exp\left(Cu^{\frac{2}{m}}\cos(\frac{\pi}{m})\right)\sin\left(Cu^{\frac{2}{m}}\sin(\frac{\pi}{m})\right)\cos(uy)du$$ Now I can not see how the CCDF is directly proportional to $C$ (I need help to understand this proportionality). Further, there is a negative sign in the expression for the CCDF. This means that the integral is also negative always? (or there is a mistake in my calculation?) I will be very grateful for your help. Thanks in advance.