<s>[INST] minimum value of $I(l)$ in definite integration If $\displaystyle I(l)=\int^{\infty}_{0}\frac{x^l}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx.$ for $0<l<5.$
Then value of $l$ for which $I(l)$ is minimum
What i tried
Put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$\displaystyle I(l)=\int^{\infty}_{0}\frac{t^{4-l}}{2t^6+4t^5+3t^4+5t^3+3t^2+4t+2}dt=\int^{\infty}_{0}\frac{x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$
$\displaystyle 2I(l)=\int^{\infty}_{0}\frac{x^l+x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$
How do i solve it.Help me 
Thanks in advance 
 [/INST] Denote $P(x)$ the polynomial of denominator, $$I'(l)=\int_0^\infty\frac{x^l\ln x}{P(x)}dx\\
=\int_0^1\frac{x^l\ln x}{P(x)}dx+\int_1^\infty\frac{x^l\ln x}{P(x)}dx\\
=\int_0^1\frac{x^l\ln x}{P(x)}dx-\int_0^1\frac{x^{4-l}\ln x}{P(x)}dx\text{ (Sub $x\mapsto 1/x$)}\\
=\int_0^1\frac{(x^l-x^{4-l})\ln x}{P(x)}dx$$
$x^l-x^{4-l}>0$ if $0<l<2$,
$x^l-x^{4-l}<0$ if $2<l<5$,
$x^l-x^{4-l}=0$ if $l=2$.
Notice $\frac{\ln x}{P(x)}<0$ if $0<x<1$, can you continue?
Can you prove it is valid to differentiate under the integral sign?
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