Suppose that π½ > 0 and that β β π [βπ½, π½]. 1) If β is even, show that t β« h(integration limit from [βπ½, π½]) = 2 β« h(integration limit from [0, π½])
We should prove that for even function h
"\\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:
"\\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
"\\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 "-\\int\\limits_{\\beta}^{0} h(x)\\,dx = \\int\\limits_{0}^{\\beta} h(x)\\,dx" .
Therefore, "\\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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