Lambert W integral October 17, 2022 Tolaso Let be a real number. Prove that Solution We have successively: Read more They might interest you:FactorizationDouble series with least common multipleζ(3) tail in floor functionA squared trigamma seriesOn the zero functionTrigonometry with golden ratioA Riemann sum IIIRoot inequalityAn infinite product
be a real number. Prove that![\[\int_0^{\infty}W_0(x^{-\alpha})\, \mathrm{d}x =\alpha^{1-\frac{1}{\alpha}} \Gamma \left(1-\frac{1}{\alpha} \right)\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-4f67b3c1ae1f9099c2d4b6b936375c6b_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \int_{0}^{\infty} W_0\left ( x^{-\alpha} \right )\, \mathrm{d}x & \overset{t =W_0 \left ( x^{-\alpha} \right )}{=\! =\! =\! =\! =\! =\! =\! =\!} \frac{1}{\alpha} \int_{0}^{\infty} \left ( t+1 \right ) t^{-1/\alpha} e^{-t/\alpha}\, \mathrm{d}t \\ &\!\!\!\!\!\!\!\overset{t = \alpha s}{=\! =\! =\! =\! =\!} \alpha^{-1/\alpha} \int_{0}^{\infty} \left ( 1+ \alpha s \right ) s^{-1/\alpha} e^{-s}\, \mathrm{d}s \\ &= \alpha^{-1/\alpha} \left ( \int_{0}^{\infty} \alpha^{-1/\alpha} e^{-s}\, \mathrm{d}s + \alpha \int_{0}^{\infty} s^{1-1/\alpha} e^{-s} \, \mathrm{d}s \right ) \\ &= \alpha^{-1/\alpha} \left [ \Gamma \left ( 1 - \frac{1}{\alpha} \right ) + \alpha \Gamma \left ( 2 - \frac{1}{\alpha} \right ) \right ] \\ &= \alpha^{-1/\alpha} \left [ \Gamma \left ( 1 - \frac{1}{\alpha} \right ) + \alpha \left ( 1 - \frac{1}{\alpha} \right ) \Gamma \left ( 1 - \frac{1}{\alpha} \right ) \right ] \\ &= \alpha^{1 - 1/\alpha} \Gamma \left ( 1 - \frac{1}{\alpha} \right ) \end{align*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-c4769e7de729b8fca7ca743d347979b7_l3.png "Rendered by QuickLaTeX.com")
be a real number. Prove that