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I'm in highschool learning about integrals and I wanted to find out what

$$ \int{x}^{x}\,dx $$

equals. I was wondering if the equation I made below is actually equal to it.

$$ \sum_{k=1}^{\infty} \frac{x^{k} \left( \sum_{m=2}^{k+1} \left( \cos(\pi m) \left( k^{k-m} \cdot \frac{k!}{(k-m+1)!} \right) \ln(x)^{k-m+1} \right) \right)}{k! \cdot k^{k-1}} $$

From what I could tell they look equal to each other when I plug them into a calculator, but I was wondering if there is somebody who is actually smart enough to be able to tell if they are.

This took me a while to come up with as I had never learned about summations/series before, and I had to teach myself the basics.

Thank you.

AllCatsAreGrey
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  • This is similar to Finding $\int x^xdx$. You also can simplify the sums, like by canceling $k!$ and writing $\cos(\pi m)=(-1)^m$ – Тyma Gaidash Sep 17 '24 at 18:34
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    $\cos(\pi m)=(-1)^m,$ not sure why you'd choose to invoke $\cos,$ whchseems to add complexity. Also, can't you cancel $k!$ and $k^{k-1}$ from the numerator and denominator to get a slightly simpler formula? – Thomas Andrews Sep 17 '24 at 18:35
  • This is a really impressive question for a first-time calculus student. Note that you can use the words "formula" or "expression" to describe what you wrote, and use the word "equation" when there's an actual equals sign. – David K Sep 18 '24 at 14:09

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Yes, but there are much simpler ways of expressing the integral in terms of series. Even your current expression can still be simplified. One way you can easily put this in terms of a series is using the well known series for $e^x$ and then substituting: $$\int x^xdx=\int e^{\ln x^x}dx=\int\sum_{k=0}^{\infty}\frac{x^k\ln^kx}{k!}dx$$

AllCatsAreGrey
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