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I'm trying to find this integral: $$\int x^\sqrt x \, dx $$

  • Wolframalpha gave me an integral. (So it does exist)
  • I tried integration by parts & tried converting it to $$ e^{\sqrt x \ln(x)} $$ then expanding $e$ by its summation notation.
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      $\begingroup$ I suppose the question is not what the integral is, but more how to derive it? $\endgroup$ Mar 16 '16 at 12:13
    • $\begingroup$ @vrugtehagel Yes! Exactly. $\endgroup$
      – Mario
      Mar 16 '16 at 12:19
    • $\begingroup$ Wolfram didn't give you an integral. It doesn't exist in terms of elementary functions. It may have given you the result for a definite one. $\endgroup$
      – Turing
      Mar 16 '16 at 12:21
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      $\begingroup$ @vrugtehagel So why the derivative isn't the same? wolframalpha.com/input/… There is a bug in wolfram online calculator. I have mathematica 10.02 on my computer and it spits out nothing. $\endgroup$
      – Turing
      Mar 16 '16 at 12:29
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      $\begingroup$ This looks like a wolfram bug - it is treating the $\sqrt{x}$ exponent like a constant. i.e. it is using the result:$$\int x^{\sqrt{a}}dx=\frac{x^{\sqrt{a}+1}}{\sqrt{a}+1}+C$$with $a=x$ $\endgroup$
      – Mufasa
      Mar 16 '16 at 12:32
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    $$x^{\sqrt{x}} = e^{\sqrt{x}\ln(x)} = \sum_{k = 0}^{+\infty} \frac{(\sqrt{x}\ln(x))^k}{k!}$$

    Thence you get

    $$\sum_{k = 0}^{+\infty} \frac{1}{k!} \int \left(\sqrt{x}\ln(x)\right)^k\ \text{d}x$$

    A repeated integration by parts gives:

    $$\int \left(\sqrt{x}\ln(x)\right)^k\ \text{d}x = \Gamma\left[1 + \frac{k}{2},\ -\left(1 + \frac{k}{2}\right)\ln(x)\right]\ln^{1 + k/2}(x)\left(-\left(1 + \frac{k}{2}\right)\ln(x)\right)^{-1 - k/2}$$

    So in the end we have

    $$\sum_{k = 0}^{+\infty} \frac{1}{k!}\left(\Gamma\left[1 + \frac{k}{2},\ -\left(1 + \frac{k}{2}\right)\ln(x)\right]\ln^{1 + k/2}(x)\left(-\left(1 + \frac{k}{2}\right)\ln(x)\right)^{-1 - k/2}\right)$$

    Gamma Function

    More here about the Gamma function

    Incomplete Gamma Function

    More here about the incomplete Gamma Function (which is the used one)

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      $\begingroup$ (+1) I think this the best we can get here! I don't know but maybe Lambert function may come in use in this problem. :) $\endgroup$ Mar 16 '16 at 12:35
    • $\begingroup$ Thank you!! Well I did the same when I had to integrate $x^x$, but in that case it was quite easier, also because of the well known Sophomore's dream. I wonder if we may find a sort of convergence to some number too here! $\endgroup$
      – Turing
      Mar 16 '16 at 12:37
    • $\begingroup$ Thanks a lot for the help! $\endgroup$
      – Mario
      Mar 16 '16 at 12:42
    • $\begingroup$ But how did you write k=0 in $$\sqrt x \ln(x)$$ ? (It'll be $$0*-\infty $$) $\endgroup$
      – Mario
      Mar 16 '16 at 12:47
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      $\begingroup$ @Mario you can see it in the limit way: $$\lim_{x\to 0} \sqrt{x}\ln(x)= 0$$ $\endgroup$
      – Turing
      Mar 16 '16 at 13:05

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