Undergrad Relation Between Beta and Gamma functions

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The discussion centers on understanding a specific proof involving the relationship between the Beta and Gamma functions. The key point is the integral expression, which states that the integral from 0 to infinity of e^{-x(1+y)} multiplied by x raised to the power of (m+n-1) equals the Gamma function divided by (1+y) raised to the power of (m+n). Participants are seeking clarification on how to apply the relevant formulas to derive this result. The conversation emphasizes the need for a step-by-step explanation of the proof to grasp the underlying concepts. Understanding this relationship is crucial for further studies in advanced mathematics.
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Proof of Beta Gamma function relation
Screenshot 2024-01-02 173019.png

So, my teacher showed me this proof and unfortunately it is vacation now. I don't understand what just happened in the marked line. Can someone please explain?
 
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The line asserts that <br /> \int_0^\infty e^{-x(1+ y)}x^{m+n-1}\,dx = \frac{\Gamma(m + n)}{(1 + y)^{m + n}}. How would you apply the given formulae to arrive at that?
 

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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