A calculus problem by Aaron Jerry Ninan

Calculus Level 4

∫ 0 ∞ ( 2 x 2 + 1 ) e − x 2 d x \large \int_{0}^{\infty }(2x^2 +1)e^{-x^{2}} \, dx

Find the value of closed form of the integral above to 2 decimal places.


The answer is 1.77.

Aaron Jerry Ninan by Aaron Jerry Ninan , 20, India

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1 solution

Relevant wiki: Gamma Function

I = ∫ 0 ∞ ( 2 x 2 + 1 ) e − x 2 d x = ∫ 0 ∞ 2 x 2 e − x 2 d x + ∫ 0 ∞ e − x 2 d x = 2 ∫ 0 ∞ x 2 × 3 2 − 1 e − x 2 d x + 2 2 ∫ 0 ∞ x 2 × 1 2 − 1 e − x 2 d x = Γ ( 3 2 ) + 1 2 Γ ( 1 2 ) Γ ( z ) is gamma function. = 1 2 Γ ( 1 2 ) + 1 2 Γ ( 1 2 ) = Γ ( 1 2 ) = π ≈ 1.77 \begin{aligned} I & = \int_0^\infty (2x^2+1) e^{-x^2} \ dx \\ & = \int_0^\infty 2x^2e^{-x^2} \ dx + \int_0^\infty e^{-x^2} \ dx \\ & = 2 \int_0^\infty x^{2\times {\color{#D61F06}\frac 32} - 1} e^{-x^2} \ dx + \frac 22 \int_0^\infty x^{2\times {\color{#D61F06}\frac 12} - 1} e^{-x^2} \ dx \\ & = \Gamma \left({\color{#D61F06}\frac 32}\right) + \frac 12 \Gamma \left({\color{#D61F06}\frac 12}\right) & \small \color{#3D99F6} \Gamma (z) \text{ is gamma function.} \\ & = \frac 12 \Gamma \left(\frac 12\right) + \frac 12 \Gamma \left(\frac 12\right) \\ & = \Gamma \left(\frac 12\right) = \sqrt \pi \approx \boxed{1.77} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Can we solve this problem with elementary methods.I think there exists a solution which uses only elementary methods.

Aaron Jerry Ninan - 4 years, 4 months ago

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Sorry, not that I know of.

Chew-Seong Cheong - 4 years, 4 months ago

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