summation of integration

Calculus Level 5

f ( m ) = ∫ − 1 1 1 − x 2 1 + m − x d x \large f(m) = \int_{-1}^{1}\frac{\sqrt{1-x^2}}{\sqrt{1+m} - x}dx

Let f ( m ) f(m) be defined as above and ∑ m = 0 99 f ( m ) = k π \displaystyle \sum_{m=0}^{99} f(m) = k \pi . Find k k

20 100 0 10

Rakshit Joshi by Rakshit Joshi , 26, India

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2 solutions

Mark Hennings
Nov 5, 2017

We have f ( m ) = ∫ − 1 1 1 − x 2 1 + m − x d x = ∫ − 1 1 1 − x 2 ( 1 + m + x ) 1 + m − x 2 d x = 1 + m ∫ − 1 1 1 − x 2 1 + m − x 2 d x = m + 1 ∫ − 1 2 π 1 2 π cos ⁡ 2 θ m + cos ⁡ 2 θ d θ = π m + 1 − m m + 1 ∫ − 1 2 π 1 2 π d θ m + cos ⁡ 2 θ = π m + 1 − 2 m m + 1 ∫ − 1 2 π 1 2 π d θ 2 m + 1 + cos ⁡ 2 θ = π m + 1 − m m + 1 ∫ − π π d ϕ 2 m + 1 + cos ⁡ ϕ = π m + 1 − m m + 1 ∫ ∣ z ∣ = 1 2 z + 2 ( 2 m + 1 ) + z − 1 d z i z = π m + 1 − 2 m m + 1 i ∫ ∣ z ∣ = 1 d z z 2 + 2 ( 2 m + 1 ) z + 1 = π m + 1 − 4 π m m + 1 R e s z = − ( 2 m + 1 ) + 2 m ( m + 1 ) ( 1 z 2 + 2 ( 2 m + 1 ) z + 1 ) = π m + 1 − 4 π m m + 1 × 1 4 m ( m + 1 ) = π m + 1 − π m \begin{aligned} f(m) & = \; \int_{-1}^1 \frac{\sqrt{1-x^2}}{\sqrt{1+m}-x}\,dx \; = \; \int_{-1}^1 \frac{\sqrt{1-x^2}(\sqrt{1+m} + x)}{1 + m - x^2}\,dx \; = \; \sqrt{1+m}\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+m - x^2}\, dx \\ & = \; \sqrt{m+1}\int_{-\frac12\pi}^{\frac12\pi} \frac{\cos^2\theta}{m + \cos^2\theta}\,d\theta \; = \; \pi\sqrt{m+1} - m\sqrt{m+1}\int_{-\frac12\pi}^{\frac12\pi}\frac{d\theta}{m + \cos^2\theta} \\ & = \; \pi\sqrt{m+1} - 2m\sqrt{m+1}\int_{-\frac12\pi}^{\frac12\pi}\frac{d\theta}{2m + 1 + \cos2\theta} \; = \; \pi\sqrt{m+1} - m\sqrt{m+1}\int_{-\pi}^\pi\frac{d\phi}{2m+1 + \cos\phi} \\ & = \; \pi\sqrt{m+1} - m\sqrt{m+1}\int_{|z|=1} \frac{2}{z + 2(2m+1) + z^{-1}} \, \frac{dz}{iz} \; = \; \pi\sqrt{m+1} - \frac{2m\sqrt{m+1}}{i}\int_{|z|=1}\frac{dz}{z^2 + 2(2m+1)z + 1} \\ & = \; \pi\sqrt{m+1} - 4\pi m\sqrt{m+1}\,\mathrm{Res}_{z = -(2m+1) + 2\sqrt{m(m+1)}} \left(\frac{1}{z^2 + 2(2m+1)z + 1}\right) \; = \; \pi\sqrt{m+1} - 4\pi m \sqrt{m+1} \times \frac{1}{4\sqrt{m(m+1)}} \\ & = \; \pi\sqrt{m+1} - \pi\sqrt{m} \end{aligned} which means that ∑ m = 0 99 f ( m ) = π 100 − π 0 = 10 π \sum_{m=0}^{99} f(m) \; = \; \pi\sqrt{100} - \pi\sqrt{0} \; = \; \boxed{10\pi}

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Vijay Simha
Nov 3, 2017

When we integrate the first integral we get pi*(sqrt(m+1) - sqrt(m))

When we sum this using the summation from m=0 to 99, we get a Telescoping Sum which collapses to pi(sqrt(100) - sqrt(0)) = 10pi

3 Helpful 0 Interesting 0 Brilliant 0 Confused

@Vijay Simha , can you please elaborate your first step, I didn't got that.

Aman Joshi - 3 years, 7 months ago

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