Answer to Question #177952 in Real Analysis for Izza Eeman

Question #177952

Suppose that 𝛽 > 0 and that β„Ž ∈ 𝑅[βˆ’π›½, 𝛽]. 1) If β„Ž is even, show that t ∫ h(integration limit from [βˆ’π›½, 𝛽]) = 2 ∫ h(integration limit from [0, 𝛽])


1
Expert's answer
2021-04-15T07:39:34-0400

We should prove that for even function h

βˆ«βˆ’Ξ²Ξ²h(t) dt=2∫0Ξ²h(t) dt.\int\limits_{-\beta}^{\beta} h(t)\,dt= 2\int \limits_{0}^{\beta} h(t)\,dt.

We know that h(-x)=h(x).


Let us divide the integral into two parts:

βˆ«βˆ’Ξ²Ξ²h(t) dt=βˆ«βˆ’Ξ²0h(t) dt+∫0Ξ²h(t) dt\int\limits_{-\beta}^{\beta} h(t)\,dt= \int\limits_{-\beta}^{0} h(t)\,dt + \int\limits_{0}^{\beta} h(t)\,dt .

We may see that the first integral may be transformed

βˆ«βˆ’Ξ²0h(t) dt=∣x=βˆ’t∣=βˆ’βˆ«Ξ²0h(βˆ’x) dx=βˆ’βˆ«Ξ²0h(x) dx\int\limits_{-\beta}^{0} h(t)\,dt = \Big| x = -t \Big| = -\int\limits_{\beta}^{0} h(-x)\,dx = -\int\limits_{\beta}^{0} h(x)\,dx because h(-x)=h(x).

And βˆ’βˆ«Ξ²0h(x) dx=∫0Ξ²h(x) dx-\int\limits_{\beta}^{0} h(x)\,dx = \int\limits_{0}^{\beta} h(x)\,dx .

Therefore, βˆ«βˆ’Ξ²Ξ²h(t) dt=∫0Ξ²h(x) dx+∫0Ξ²h(t) dt=2∫0Ξ²h(t) dt\int\limits_{-\beta}^{\beta} h(t)\,dt= \int\limits_{0}^{\beta} h(x)\,dx + \int\limits_{0}^{\beta} h(t)\,dt = 2\int\limits_{0}^{\beta} h(t)\,dt .


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