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Definite integral challenge...

DreamWeaver

Well-known member
Sep 16, 2013
337
For \(\displaystyle m \in \mathbb{Z}^+\), and \(\displaystyle a, \, z \in \mathbb{R} > 0\), evaluate the definite integral:


\(\displaystyle \int_0^z\frac{x^m}{(a+\log x)}\,dx\)





[I'll be adding a few generalized forms like this in the logarithmic integrals thread, in Maths Notes, shortly... (Heidy) ]
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
You can find an antiderivative in terms of the exponential integral.


$ \displaystyle \int \frac{x^{m}}{a+\ln x} \ dx = \int \frac{e^{u(m+1)}}{a+u} \ du = e^{-a(m+1)} \int \frac{e^{m(w+1)}}{w} \ dw $

$ \displaystyle = e^{-a(m+1)} \ \text{Ei} \Big( w(m+1) \Big) + C $

$ \displaystyle = e^{-a(m+1)} \ \text{Ei} \Big( (a+u) (m+1) \Big) + C $

$ \displaystyle = e^{-a(m+1)} \ \text{Ei} \Big( (a+\ln x) (m+1) \Big) + C $


And I think you need more restrictions on the parameters to guarantee convergence.