If $\displaystyle f(a)=\int^{\infty}_{0}\frac{x^a}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}$is minimum.Then real value of $a$ is
Try: $$f(a)=\int^{\infty}_{0}\frac{x^{a-3}}{2(x^6+x^{-6})+4(x^2+x^{-2})+3(x+x^{-1})+5}dx$$
given $x>0$. So using A.M$\geq$ G.M,
$x^k+x^{-k}\geq 2$ for $k=1,2,6$
Could some help me to solve it, Thanks
$$f'(a)=\int_{0}^{+\infty}\frac{x^a\log(x)}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}\,dx $$ equals $$ f'(a)=\int_{0}^{+\infty}\frac{x^{a-3}\log(x)}{2(x^3+x^{-3})+4(x^2+x^{-2})+3(x+x^{-1})+5}\,dx $$ or $$ f'(a)=\int_{0}^{1}\frac{(x^{a-3}-x^{1-a})\log(x)}{2(x^3+x^{-3})+4(x^2+x^{-2})+3(x+x^{-1})+5}\,dx $$ so $f'(a)$ clearly equals zero when $a-3=1-a$, i.e. for $a=2$. Due to the substitution $x\mapsto\frac{1}{x}$ we have $f(2-u)=f(2+u)$ and now it is not difficult to show that the minimum of $f$ is indeed $f(2)\approx 0.06822$, since $f$ is (log)convex.