For $\alpha>0$, consider the integral $$ \int_{0}^{\infty}\exp\left ( \frac{1}{2}x-x^\alpha \right )\,\mathrm{d}x $$ For which values of $\alpha$, does this integral converge?
I made lots of trying to conclude that this integral converges for $\alpha\geq 1$. The case $\alpha=1$ is easy, so we check $\alpha>1$. Call the integrand $f_\alpha(x)$. Then $f_\alpha'(x)\leq 0$ for $x\geq 1/(2\alpha)^{\alpha-1}$. Put $m_\alpha:=1/(2\alpha)^{\alpha-1}$. It is reduced to check if the following integral converges $$ \int_{m_\alpha}^{\infty}\exp\left ( \frac{1}{2}x-x^\alpha \right )\,\mathrm{d}x $$ because $\int_{0}^{m_\alpha}\exp\left ( \frac{1}{2}x-x^\alpha \right )\,\mathrm{d}x$ is finite. We know that $x\mapsto 1/f_\alpha(x)=\exp(x^\alpha-1/2x)$ is increasing for $x\geq 1/(2\alpha)^{\alpha-1}$. So, I desire to find the lower bound of $1/f_\alpha$. By means of Taylor series of $\exp$ on $\mathbb{R}$, we get $1+x^\alpha-1/2x\leq 1/f_\alpha(x)$ for all $x\in \mathbb{R}$, if I am not mistaken. Then we get $f_\alpha(x)\leq 1/(1+x^\alpha-1/2x)$ for all $x\in \mathbb{R}$. So, we should find out if the integral of $1/(1+x^\alpha-1/2x)$ converges in $[m_\alpha,\infty)$ to conclude. Am I on right on this track? Is there a simpler way? Like, to find a suitable substitution to convert this integral in terms of the Euler's Gamma function?
Yes. A simpler way is to note that for $x\geq 1$ and $\alpha\geq 1$ $$\exp({1\over 2}x-x^\alpha)\le \exp(-{x^\alpha\over 2})\le \exp(-{x\over 2})$$