Integrating Complex Exponentials

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The integral of the complex exponential e^(ix) from 0 to infinity is evaluated to be -1/i, with the upper limit contributing zero due to the oscillatory nature of the function. This behavior arises because the oscillations cancel out over the infinite interval, leading to a converging result. The discussion highlights the importance of understanding complex analysis in evaluating such integrals. The conclusion reached demonstrates a successful proof of the integral's value. This topic underscores the intriguing properties of complex exponentials in mathematical analysis.
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The integral of a complex exponential ( e^(ix) ) over x from 0 to infinity is supposedly such that the value of the definite integral at the upper limit is zero and so it's just -1/i. Why is this? It's just an oscillating function after all.
 
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Nevermind, I figured it out a way to prove it.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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