What is the unique special function in this integral problem?

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SUMMARY

The integral problem discussed involves the evaluation of the integral $$\int_0^{\infty}\frac{t^{\frac{m+1}{n}-1}}{1+t}dt$$ which is equated to the beta function $\mathcal{B}(\frac{m+1}{n},1-\frac{m+1}{n})$. This integral is recognized as a transformation of the ordinary Euler's Beta function, specifically with the substitution $\frac{m+1}{n} \equiv \alpha$. The discussion highlights the method of changing variables to simplify the integral, ultimately reverting to the original parameters $m$ and $n$.

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  • Understanding of integral calculus, specifically improper integrals.
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Fred Wright
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While studying the solution to a integral problem I found online I ran across a special function I am unfamiliar with. The integral is
$$
\int_0^{\infty}\frac{t^{\frac{m+1}{n}-1}}{1+t}dt=\mathcal{B}(\frac{m+1}{n},1-\frac{m+1}{n})
$$
This certainly isn't the normal beta function. What is it? Can anyone direct me to a reference on this function?
 
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It's just the "normal Euler's Beta function" in the RHS but with "intimidating arguments" :smile:. One trick is to simply replace ##\frac{m+1}{n} \equiv \alpha## in your exponent under the integral sign, do the integral and revert to ##m,n## at the end.
 
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\int_0^1 t^{p-1}(1-t)^{q-1}\,dt = \int_0^\infty \frac{z^{p-1}}{(z + 1)^{p+q}}\,dz where z = t/(1-t).
 
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