# Integral of y = x^x

The function of x raised to the power x looks simple enough to be integrated, but unfortunately, it falls into the category of non-integrable functions. In other words, you cannot find the antiderivative of f(x) = x^x in terms of elementary functions.

However, there are many ways to approximate the antiderivative of x^x using infinite series and asymptotic series, which allow you to estimate the area under a segment of the curve to any degree of accuracy depending on how many terms you carry. You can also estimate the area under f(x) = x^x with Riemann sums, using left endpoints, right endpoints, the trapezoid average, or midpoints. Below are several techniques for estimating the antiderivative of x^x along with some example computations.

## Graph of y = x^x

## How to Integrate x^x Approximately With an Infinite Series

Convergent and asymptotic infinite series are useful in finding approximate antiderivatives. The Taylor series expansion of f(x) = x^x centered at the point x = 1 is

f(x) =

1 + (x-1) + (x-1)^{2} + (x-1)^{3}/2 +

(x-1)^{4}/3 + (x-1)^{5}/12 + 3(x-1)^{6}/40 +

(x-1)^{7}/120 + 59(x-1)^{8}/2520

+ 71(x-1)^{9}/5040 + ...

Integrating this term by term, you get

∫ x^x dx =

x - (x-1)^{2}/2 + (x-1)^{3}/3 + (x-1)^{4}/8 +

(x-1)^{5}/15 + (x-1)^{6}/60 + 3(x-1)^{7}/280 +

(x-1)^{8}/960 + 59(x-1)^{9}/22680 + 71(x-1)^{10}/50400 + ...

Unfortunately, this power series is not everywhere-convergent and works best for values of x in the interval (0, 2).

## Another Series Antiderivative of x^x

Using the fact that x^x = e^(x*Ln(x)) you can come up with another series for the antiderivative of x^x. First we use the Taylor series expansion of e^w:

e^w = 1 + w + w^{2}/2! + w^{3}/3! + w^{4}/4! + w^{5}/5! + ...

Replacing w with x*ln(x) gives us

e^(x*Ln(x)) = 1 + xLn(x) + [xLn(x)]^{2}/2! + [xLn(x)]^{3}/3! + [xLn(x)]^{4}/4! + ...

Now we integrate this series term by term using the the technique of integration by parts. Here are the first few terms of the antiderivative:

∫ e^(x*Ln(x)) dx =

x + (x^{2}/4)[2*Ln(x) - 1] + (x^{3}/54)[9*Ln(x)^2 - 6*Ln(x) + 2] +

(x^{4}/768)[32*Ln(x)^3 - 24*Ln(x)^2 + 12*Ln(x) - 3] + ...

Antiderivatives of functions of the form g(x) = (x*Ln(x))^n get hairy and unwieldy as n increases, as you can see in the series above. Although this series in convergent, it converges too slowly to be of practical use for large values of x.

## Estimating the Integral of x^x With Riemann Sums

A more efficient way to compute definite integrals of f(x) = x^x is with Riemann sums, a sampling method based on approximating the area under a curve with rectangular strips. The larger the number of subintervals you use, the more accurate the approximation. As an example, let's approximate the area under x^x from x = 0 to x = 1 using 10 sub-intervals and a left Riemann sum, as shown in the image below.

The sum of the areas of these 10 rectangular strips is

0.1 * [0^0 + 0.1^0.1 + 0.2^0.2 + 0.3^0.3 + 0.4^0.4 +

0.5^0.5 + 0.6^0.6 + 0.7^0.7 + 0.8^0.8 + 0.9^0.9]

≈ 0.787733

Here we use the fact that 0^0 = 1 since lim(x→0) x^x = 1. If we use the same number of sub-intervals but us a right Riemann sum, the estimated area is

0.1 * [0.1^0.1 + 0.2^0.2 + 0.3^0.3 + 0.4^0.4 + 0.5^0.5 +

0.6^0.6 + 0.7^0.7 + 0.8^0.8 + 0.9^0.9 + 1^1]

≈ 0.787733

which is exactly equal to the left Riemann sum. Increasing the number of sub-intervals gives us a better estimate of the area under the curve x^x from 0 to 1. If we use n = 1,000,000,000 we get an area of 0.783431.

## Some Definite Integrals

Here are several definite integrals involving the function x^x and x^(-x). Unlike special non-integrable functions, these definite integrals cannot be represented in terms of pi or natural logarithms of numbers. These were calculated using a numerical integrator, such as the one programmed into graphing calculators.

## Comments

So there is no antiderivatve formula for x^x? How do you find the exact area in that case, not the approximate area? I remember in AP calculus we learned about functions that do not have antiderivatives, e^x^2 and sqrt(x^4 + 1). How do you know when you have an integral that can't be solved?