A calculus problem by Amal Hari

Integrating by parts,

$I_{999}=(-t^{999}e^{-t}+999\int t^{998}e^{-t}dt)^{∞}_{0}$

After putting the limits we get this as,

$I_{999}=999I_{998}$

If we follow the same procedure again, we will get that $I_{n}=nI_{n-1}$ . Thus,

$I_{999}=999×998×997....×2×1×(\int ^{∞}_{0} e^{-t}dt)=999!$

$\begin{aligned} I & = \int_0^\infty e^{-t}t^{999} dt & \small \color{#3D99F6} \text{Since gamma function }\Gamma (s) = \int_0^\infty t^{s-1}e^{-t} dt \\ & = \Gamma (1000) & \small \color{#3D99F6} \text{and } \Gamma(n) = (n-1)! \\ & = 999! \end{aligned}$

Therefore, $x = \boxed{999}$ .

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Reference: Gamma function