We should prove that for even function h
β« β Ξ² Ξ² h ( t ) βd t = 2 β« 0 Ξ² h ( t ) βd t . \int\limits_{-\beta}^{\beta} h(t)\,dt= 2\int \limits_{0}^{\beta} h(t)\,dt. β Ξ² β« Ξ² β h ( t ) d t = 2 0 β« Ξ² β h ( t ) d t .
We know that h(-x)=h(x).
Let us divide the integral into two parts:
β« β Ξ² Ξ² h ( t ) βd t = β« β Ξ² 0 h ( t ) βd t + β« 0 Ξ² h ( t ) βd t \int\limits_{-\beta}^{\beta} h(t)\,dt= \int\limits_{-\beta}^{0} h(t)\,dt + \int\limits_{0}^{\beta} h(t)\,dt β Ξ² β« Ξ² β h ( t ) d t = β Ξ² β« 0 β h ( t ) d t + 0 β« Ξ² β h ( t ) d t .
We may see that the first integral may be transformed
β« β Ξ² 0 h ( t ) βd t = β£ x = β t β£ = β β« Ξ² 0 h ( β x ) βd x = β β« Ξ² 0 h ( x ) βd x \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 β Ξ² β« 0 β h ( t ) d t = β£ β£ β x = β t β£ β£ β = β Ξ² β« 0 β h ( β x ) d x = β Ξ² β« 0 β h ( x ) d x because h(-x)=h(x).
And β β« Ξ² 0 h ( x ) βd x = β« 0 Ξ² h ( x ) βd x -\int\limits_{\beta}^{0} h(x)\,dx = \int\limits_{0}^{\beta} h(x)\,dx β Ξ² β« 0 β h ( x ) d x = 0 β« Ξ² β h ( x ) d x .
Therefore, β« β Ξ² Ξ² h ( t ) βd t = β« 0 Ξ² h ( x ) βd x + β« 0 Ξ² h ( t ) βd t = 2 β« 0 Ξ² h ( t ) βd t \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 β Ξ² β« Ξ² β h ( t ) d t = 0 β« Ξ² β h ( x ) d x + 0 β« Ξ² β h ( t ) d t = 2 0 β« Ξ² β h ( t ) d t .
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