Definite Integral from 0 to 1 of Difference of Powers of x over Logarithm of x

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Theorem

$\ds \int_0^1 \frac {x^m - x^n} {\ln x} \rd x = \map \ln {\frac {m + 1} {n + 1} }$

where $m$ and $n$ are real numbers with $m, n > -1$.


Proof

Let:

$x = e^{-u}$

We have, by Derivative of Exponential Function:

$\dfrac {\d x} {\d u} = -e^{-u}$

By Exponential Tends to Zero and Infinity:

as $x \to 0$, $u \to \infty$

By Exponential of Zero:

as $x \to 1$, $u \to 0$.

So:

\(\ds \int_0^1 \frac {x^m - x^n} {\ln x} \rd x\) \(=\) \(\ds -\int_\infty^0 e^{-u} \frac {e^{-m u} - e^{-n u} } {\ln e^{-u} } \rd u\) substituting $x = e^{-u}$
\(\ds \) \(=\) \(\ds -\int_0^\infty \frac {e^{-\paren {m + 1} u} - e^{-\paren {n + 1} u} } u \rd u\) Reversal of Limits of Definite Integral, Exponential of Sum
\(\ds \) \(=\) \(\ds \int_0^\infty \frac {e^{-\paren {n + 1} u} - e^{-\paren {m + 1} u} } u \rd u\)
\(\ds \) \(=\) \(\ds \map \ln {\frac {m + 1} {n + 1} }\) Definite Integral to Infinity of $\dfrac {e^{-a x} - e^{-b x} } x$

$\blacksquare$


Sources

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