nice analysis exercise asked by Sourav De in the math group and edited by me

Exercise:

Consider an even function $f$ and $\alpha ,\beta $ ; $\alpha <\beta $ be two roots for $g\left( x \right)=a{{x}^{2}}+bx+c$

Show that, $\int_{\alpha }^{\beta }{\frac{{{e}^{f\left( \frac{g\left( x \right)}{x-\alpha } \right)}}}{{{e}^{f\left( \frac{g\left( x \right)}{x-\alpha } \right)}}+{{e}^{f\left( \frac{g\left( x \right)}{x-\beta } \right)}}}dx}=\frac{\sqrt{\Delta }}{2a}\,\,\,\,\,,\,\,\Delta ={{b}^{2}}-4ac$

Solution: we have $\alpha ,\beta $ are roots for $g\left( x \right)$ so $x-\alpha \,\And \,\,x-\beta $ are factors

So $g\left( x \right)=a\left( {{x}^{2}}+\frac{b}{a}x+\frac{c}{a} \right)=a\left( {{x}^{2}}-Sx+p \right)=a\left( {{x}^{2}}-\left( \alpha +\beta  \right)x+\alpha \beta  \right)=a\left( x-\alpha  \right)\left( x-\beta  \right)$

Notice that $f$ is even function then $f\left( -x \right)=f\left( x \right)$  symmetric w.r.t y-axis

So $f\left( \frac{g\left( x \right)}{x-\alpha } \right)=f\left( \frac{a\left( x-\alpha  \right)\left( x-\beta  \right)}{x-\alpha } \right)=f\left( a\left( x-\beta  \right) \right)$ and $f\left( \frac{g\left( x \right)}{x-\beta } \right)=f\left( a\left( x-\alpha  \right) \right)$

So $I=\int_{\alpha }^{\beta }{\frac{{{e}^{f\left( \frac{g\left( x \right)}{x-\alpha } \right)}}}{{{e}^{f\left( \frac{g\left( x \right)}{x-\alpha } \right)}}+{{e}^{f\left( \frac{g\left( x \right)}{x-\beta } \right)}}}dx}=\int_{\alpha }^{\beta }{\frac{{{e}^{f\left( a\left( x-\beta  \right) \right)}}}{{{e}^{f\left( a\left( x-\beta  \right) \right)}}+{{e}^{f\left( a\left( x-\alpha  \right) \right)}}}dx}$

Using the Property $\int_{\alpha }^{\beta }{f\left( x \right)dx}=\int_{\alpha }^{\beta }{f\left( \alpha +\beta -x \right)dx}$

So $f\left( a\left( x-\alpha  \right) \right)=f\left( ax-a\alpha  \right)=f\left( a\left( \alpha +\beta -x \right)-a\alpha  \right)=f\left( a\left( \beta -x \right) \right)=f\left( -a\left( x-\beta  \right) \right)$

And $f\left( a\left( x-\beta  \right) \right)=f\left( ax-a\beta  \right)=f\left( a\left( \alpha +\beta -x \right)-a\beta  \right)=f\left( a\left( \alpha -x \right) \right)=f\left( -a\left( x-\alpha  \right) \right)$

So $I=\int_{\alpha }^{\beta }{\frac{{{e}^{f\left( a\left( x-\beta  \right) \right)}}}{{{e}^{f\left( a\left( x-\beta  \right) \right)}}+{{e}^{f\left( a\left( x-\alpha  \right) \right)}}}dx}$   as $f$ is even function

So $2I=\int_{\alpha }^{\beta }{dx}\Leftrightarrow I=\frac{\beta -\alpha }{2}$

Observe that ${{\left( \beta -\alpha  \right)}^{2}}={{\beta }^{2}}-2\alpha \beta +{{\alpha }^{2}}+4\alpha \beta -4\alpha \beta ={{\left( \alpha +\beta  \right)}^{2}}-4\alpha \beta $


$\Leftrightarrow \beta -\alpha =\sqrt{{{\left( \alpha +\beta  \right)}^{2}}-4\alpha \beta }=\sqrt{\frac{{{b}^{2}}-4ac}{{{a}^{2}}}}=\frac{1}{a}\sqrt{{{b}^{2}}-4ac}=\frac{1}{a}\sqrt{\Delta }$

Thus $I=\int_{\alpha }^{\beta }{\frac{{{e}^{f\left( \frac{g\left( x \right)}{x-\alpha } \right)}}}{{{e}^{f\left( \frac{g\left( x \right)}{x-\alpha } \right)}}+{{e}^{f\left( \frac{g\left( x \right)}{x-\beta } \right)}}}dx}=\frac{\sqrt{\Delta }}{2a}$