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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.
    • Hosein Rahnama
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    Mario
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      I suppose the question is not what the integral is, but more how to derive it? –  Mar 16 '16 at 12:13
    • @vrugtehagel Yes! Exactly. – Mario Mar 16 '16 at 12:19
    • 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. – Enrico M. Mar 16 '16 at 12:21
    • @1over137, check the wolfram output. It gives you an indefinite integral closed form. –  Mar 16 '16 at 12:22
    • @1over137 How do you know it doesn't exist? – Mario Mar 16 '16 at 12:23
    • Because there is no function $f(x)$ whose derivative is that one. And I'm going to write down a result which is not a standard function. @Mario – Enrico M. Mar 16 '16 at 12:24
    • @vrugtehagel then link us the output. – Enrico M. Mar 16 '16 at 12:24
    • The link: http://www.wolframalpha.com/input/?i=integral+x%5E(sqrt(x)) and the integral you get: $\frac{x^{\sqrt{x}+1}}{\sqrt{x}+1}+c$. You might be confused with $\int x^x$ –  Mar 16 '16 at 12:25
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      @vrugtehagel So why the derivative isn't the same? http://www.wolframalpha.com/input/?i=Derivative+(x%5E(sqrt(x)%2B1))%2F(sqrt(x)%2B1) There is a bug in wolfram online calculator. I have mathematica 10.02 on my computer and it spits out nothing. – Enrico M. Mar 16 '16 at 12:29
    • It seems that there is no answer. You may calculate only approximate answer by expanding the function. – Ali Mar 16 '16 at 12:29
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      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 $a=x$ – Mufasa Mar 16 '16 at 12:32
    • @Ali That's what I did. But I was hoping to derive that result wolfram gave. – Mario Mar 16 '16 at 12:33
    • @Mario - wolframs result is a bug - see my comment above. The actual result is much more complicated as shown by 1over below. Wolfram is known to have some bugs in it - it is not perfect :) – Mufasa Mar 16 '16 at 12:33
    • That what Wolfram is giving cannot be an answer since a derivative of $\frac{x^{\sqrt{x}+1}}{\sqrt{x}+1}$ cannot avoid having $\log$ in the result. It seems that Wolfram is treating $x$ in the power as a separate variable. –  Mar 16 '16 at 12:35
    • As you see in comments above it cannot be true it used $x$ as a constant such as $a$ in power. – Ali Mar 16 '16 at 12:36
    • WOLFRAM BUG ... the first one ever!!! (not) – GEdgar Mar 17 '16 at 15:57

    1 Answers1

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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)

    Enrico M.
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      (+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. :) – Hosein Rahnama Mar 16 '16 at 12:35
    • 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! – Enrico M. Mar 16 '16 at 12:37
    • Thanks a lot for the help! – Mario Mar 16 '16 at 12:42
    • But how did you write k=0 in $$\sqrt x \ln(x)$$ ? (It'll be $$0*-\infty $$) – Mario Mar 16 '16 at 12:47
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      @Mario you can see it in the limit way: $$\lim_{x\to 0} \sqrt{x}\ln(x)= 0$$ – Enrico M. Mar 16 '16 at 13:05