True/false convergence of integral from 1 to infinity

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SUMMARY

The integral convergence property states that if the integral \(\int^∞_1{h(x)dx}\) converges for a continuous function \(h(x)\) defined for \(x > 0\), then the integral \(\int^∞_1{h(\frac{x}{a})dx}\) also converges for any constant \(0 < a < 1\). This conclusion is based on the substitution method in calculus, which demonstrates that the behavior of the function under transformation does not affect the convergence of the integral. The assertion is confirmed as true, countering initial assumptions about the size of \(h(x/a)\) compared to \(h(x)\).

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  • Knowledge of substitution techniques in integration.
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Suppose h(x) is a continuous function for x > 0. If [tex]\int^∞_1{h(x)dx}[/tex] converges then for constant 0 < a < 1, [tex]\int^∞_1{h(\frac{x}{a})dx}[/tex] also converges.

The answer is true. Anyone care to explain why? I would have chosen false, because I was thinking that h(x/a) is larger than h(x) so we wouldn't know if it converges or not.
 
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hint: substitution :wink:
 

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