X x n x
is a specific positive integer known as a binomial coefficient.
(When an exponent is zero, the corresponding power expression is taken to be 1 and this multiplicative factor is often omitted from the term.
In don't think any further simplification is possible. $ By the series expansion of $^$: $$^ = \sum _^$$ Thus $$\int _^ dx=\sum _^=\sum _^$$ Let $u = ^ $, $dv = ^ dx $, $du = \frac dx$ and $v=\frac$, then using integration by parts, we arrive at $$\lim _ =\lim _ -\lim _ dx$$ which becomes $$\lim _ =-\int _^ ^dx = \frac$$ Therefore $$\int _^ dx=\sum _^$$ On can find a compendium of properties of the special function : $$\text(\alpha\:;\:x)=\int_0^x t^dt$$ and the particular case : $$\int x^x dx = \text(1\:;\:x) \text$$ in : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function Many people have pointed out that the integral you are looking for is equivalent to, $\sum_^ \frac \int_^x^ln(x)^ndx$ But the integral within this equation can be simplified to $\int_^x^ln(x)^ndx = \left(-1\right)^n\left(n 1\right)^\Gamma \left(n 1,\left(n 1\right)\ln \left(\frac\right)\right)$ Where $\Gamma \left(n 1,\left(n 1\right)\ln \left(\frac\right)\right) = n! Simplify the first equation, and you will get, $\int_^x^dx = \sum _^\left(\frac\right)$ A demonstration of this function may be found on the desmos graphing calculator: https:// Start from the opposite task.
If $\displaystyle \int x^x \, dx=F(x)$ then $\displaystyle F'(x)=x^x$ First we need to find asymptotic evaluation of the integral.
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The earliest known reference to this combinatorial problem is the Chandaḥśāstra by the Indian lyricist Pingala (c. Blaise Pascal studied the eponymous triangle comprehensively in the treatise Traité du triangle arithmétique (1653).