Integrals

Calculus Level 4

I k = ∫ 0 ∞ 1 1 + k x 10 d x \large \displaystyle I_k = \int_{0}^{\infty} \dfrac{1}{1+k x^{10}} \, dx

Find a closed form of the integral above. Submit your answer lim ⁡ k → ∞ I k \displaystyle \lim_{k\to\infty} I_k .


The answer is 0.

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Kushal Bose by Kushal Bose , 28, India

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1 solution

Chew-Seong Cheong
Apr 11, 2017

Relevant wiki: Beta Function

I = ∫ 0 ∞ 1 1 + k x 10 d x Let u = k x 10 , d u = 10 k x 9 d x = ∫ 0 ∞ u − 9 10 10 k 10 ( 1 + u ) d u B ( m , n ) = ∫ 0 ∞ u m − 1 ( 1 + u ) m + n d u = 1 10 k 10 B ( 1 10 , 9 10 ) where B ( m , n ) is beta function. \begin{aligned} I & = \int_0^\infty \frac 1{1+kx^{10}} dx & \small \color{#3D99F6} \text{Let }u=kx^{10}, \ du = 10kx^9 \ dx \\ & = \int_0^\infty \frac {u^{-\frac 9{10}}}{10\sqrt[10]k(1+u)} du & \small \color{#3D99F6} B(m,n) = \int_0^\infty \frac {u^{m-1}}{(1+u)^{m+n}} du \\ & = \frac 1 {10\sqrt [10] k} B \left(\frac 1{10}, \frac 9{10} \right) & \small \color{#3D99F6} \text{where } B(m,n) \text{ is beta function.} \end{aligned}

⟹ lim ⁡ k → ∞ I = 0 \displaystyle \implies \lim_{k \to \infty} I = \boxed{0}

6 Helpful 0 Interesting 2 Brilliant 0 Confused

Oh my God I thought it can written in closed form. It is clearly related to beta function. For calculating the limit we can directly apply dominated convergence theorem. Assume k>1, 1/(1+kx^10) < 1/(1+x^10). With some trivial justification we get the 1/(1+x^10) is lebesgue integrable. By applying dominated convergence theorem we get the limit=0.

Srikanth Tupurani - 2 years, 11 months ago

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