How Does the Denominator x+a Become 1-t in Integral Substitution?

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    Integral Substitution
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SUMMARY

The discussion focuses on the transformation of the denominator in the integral \(\int_0^\infty \frac{x^{\mu-1}dx}{x+a}\) when \(a<0\). By substituting \(b=-a\) and letting \(x=bt\), the denominator changes from \(x+a\) to \(1-t\) after factoring out \(-b\). This transformation is crucial for simplifying the integral for further evaluation.

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Belgium 12
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Hi members,

see attached Pdf file.
For a<0,write b=-a and let x=bt.Then

My question:

how becomes the denominator x+a to 1-t? I don't see it

Thank you
 

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The integral is
\int_0^\infty \frac{x^{\mu-1}dx}{x+ a}

You attachment says "if a< 0 let b= -a" so becomes
\int_0^\infty \frac{x^{\mu-1}dx}{x- b}

and "let x= bt". The denominator is x- b= bt- b= b(t- 1)

Factor out "-b" leave 1- t in the denominator,
 

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