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$$\int_1^{10}x^xdx$$ I figured it with a relative error of 1% and I have a response $$≈0.3*10^{10}$$ But I don´t know how to accurately calculate it ...

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You cannot find a closed form for $\int x^{x}dx$ (see… ) – Américo Tavares Nov 1 '10 at 21:36
@Tavares, you don't need to find the antiderivative of a function to accurately calculate its integral, there are many ways to find a convergence series. – Eric Haengel Nov 1 '10 at 21:39
If you just want more accuracy or to check the accuracy of any method, Wolfram alpha can help you out.^10+x^x – Dinesh Nov 1 '10 at 21:45
@Haengel, I know. My comment was an information in case OP was trying to find a closed form. – Américo Tavares Nov 1 '10 at 22:11
The problem of course with using power series is that they suck at approximating far away from the point of expansion. I would actually just suggest using numerical quadrature if one wants accurate values. – J. M. Nov 2 '10 at 0:18

One way to solve this problem is to expand in terms of a power series as follows:

First notice that

$x^x = e^{\ln(x) x} = \sum_{n=0}^\infty \frac{(\ln(x) x)^n}{n!} $

Integrating on both sides from $1$ to $10$ gives

$\int_1^{10} x^x dx = \sum_{n=0}^\infty \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx $

The function $x^n ln(x)^n$ is monotonically increasing, so the integral inside the sum is bounded by:

$ \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx < \frac{9}{n!} \cdot \ln(10)^n \cdot 10^n $

So if you want the sum to whatever precision, you can use that expression to determine how small the remainder term is. Next, how on earth do you compute that integral? Well it looks ugly but there is a recursive formula. Let $k, l$ be positive integers. Then using integration by parts

$ \int_1^{10} x^k \ln(x)^l dx = \frac{x^{k+1} ln(x)^l}{k+1} \Bigr \vert_1^{10} - \frac{l}{k+1} \int_1^{10} x^k \ln(x)^{l-1} dx $

Since you chose the limits of integration to be from $1$ to $10$ you might have to take quite a few terms in this series in order to compute an accurate value, but that is to be expected as $x^x$ blows up so quickly.

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The problem with this approach is that not only do you have to take a lot of terms (for instance, using the fact that $n! \approx (n/e)^n$, we see that the terms don't even get back to unity until n is roughly 60 or so), the terms themselves are immensely large and the negation involved in the integration by parts means they'll have to be calculated extremely precicely to navigate cancellative effects. Much better, as J.M. mentioned, to just use quadrature. – Steven Stadnicki Nov 9 '10 at 19:05

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