Can the present be absent?

Substitute $\small{\color{#D61F06}{x=\tan^2\theta}}$ such that $\small{\color{#D61F06}{\mathrm{d}x=2\tan \theta\sec^2 \theta ~\mathrm{d}\theta=\dfrac{2\sin \theta~\mathrm{d}\theta}{\cos^3\theta}}}$ so that $\left(\small{\text{Also use }\color{#69047E}{1+\tan^2\theta=\dfrac1{\cos^2\theta}}}\right)$ :

$\mathscr{E}=2\displaystyle\int_0^{\pi/2}\sin^{4031}\theta~ \underbrace{\cos \theta~\mathrm{d}\theta}_{\mathrm{d}(\sin \theta)}$

$\large =\left[\dfrac{2\sin^{4032}\theta}{4032}\right]_0^{^{\pi/2}}=\boxed{\dfrac1{2016}}$

$\Large \therefore 1+2016=\boxed{\color{#007fff}{2017}}$

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$\small{\color{grey}{\text{ALT: }}}\large\mathscr{E} = -\displaystyle \int_{0}^{\infty} \dfrac{\overbrace{\frac{-1}{x^2}}^{\mathrm{d}\left(1+\frac 1x\right)}}{\left(1+\frac 1x\right)^{2017}} \mathrm{d}x$

$\large = \left[\dfrac{\left(1+\frac 1x\right)^{-2016}}{2016}\right]_0^{\infty}=\boxed{\dfrac1{2016}}$

$\displaystyle \beta(a,b)=\int_{0}^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}}$

So our integral equals = $\displaystyle \beta(2016,1) = \frac{\Gamma(2016)\Gamma(1)}{\Gamma(2017)} = \frac{\color{#D61F06}{1}}{\color{#3D99F6}{2016}}$

So answer is : $\boxed{1+2016=2017}$

$I=\displaystyle∫_{0}^{\infty} \dfrac{x^{2015}}{(1+x)^{2017}} dx=$ $\displaystyle∫_{0}^{\infty}\dfrac{x^{2017}}{(1+x)^{2017}*x^2}=$ $\displaystyle∫_{0}^{\infty}\dfrac{1}{(\dfrac{1}{x}+1)^{2017}*x^2}dx$

Now Let $\dfrac{1}{x} = t$

$I = \displaystyle∫_{0}^{\infty} \dfrac{1}{(t+1)^{2017}}dt=$

$\displaystyle\dfrac{-1}{2016}\large \left[\dfrac{1}{(t+1)^{2016}}\right]_{0}^{\infty} =$

$\displaystyle\dfrac{-1}{2016}(0-1) = \dfrac{1}{2016}= \dfrac{a}{b}$

Therefore, $a+b = 1+2016=2017$

Also one can put z=1+(1/x)