Intergrate x^-x
To integrate x^(-x), we can use the technique of integration by parts. The formula for integration by parts is:
∫ u * dv = u * v - ∫ v * du,
where u and v are functions of x, and du and dv are their respective differentials.
Let's choose u = x^(-x) and dv = dx. Taking the differentials of both sides, we have du = (-x * x^(-x-1))dx and v = ∫dv = ∫dx = x.
Using the formula for integration by parts, we can rewrite the integral as:
∫ x^(-x)dx = ∫ u * dv = u * v - ∫ v * du
= x^(-x) * x - ∫ x * (-x * x^(-x-1))dx.
Simplifying this expression, we get:
∫ x^(-x)dx = x^(1-x) + ∫ x^(2-x)dx.
Now we need to integrate ∫ x^(2-x)dx. We can rewrite this as:
∫ x^(2-x)dx = ∫ (x^2 * x^(-x))dx = ∫ (x^2 * e^(-x * ln(x)))dx,
where we have used the fact that x^(-x) can be written as e^(-x * ln(x)).
This integral does not have a simple closed form solution and requires numerical methods or special functions to evaluate.
Therefore, the final result for the integral is:
∫ x^(-x)dx = x^(1-x) + ∫ x^(2-x)dx.
To integrate x^-x, we can use the substitution method. Let's go step by step.
Step 1: Substitute x^-x as e^(-x * ln(x)).
So, the integral becomes ∫ e^(-x * ln(x)) dx.
Step 2: Now, let's use the substitution u = -x.
Therefore, du = -dx.
Step 3: Substitute the value of du and change the limits of the integral accordingly.
When x = 0, u = 0, and when x approaches infinity, u approaches negative infinity.
So, the new integral becomes -∫ e^(u * ln(-u)) du, with the limits going from 0 to -∞.
Step 4: Simplify the integral further.
Since we have a negative sign outside the integral and the exponent is an odd function, we can use the property e^(-a) = 1/e^(a) to rewrite the integral as:
-∫ (1 / e^(u * ln(-u))) du.
Step 5: Rewrite the expression inside the integral in terms of absolute value.
Since ln(-u) is undefined for positive values of u, we take the absolute value to avoid any issues with the logarithm.
The integral becomes -∫ (1 / e^(u * ln|u|)) du.
Step 6: Evaluate the integral using the substitution method.
Let's substitute v = u * ln|u|.
Then, dv = (ln|u| + 1) du.
Step 7: Rewrite the above integral in terms of v.
The integral becomes:
-∫ (1 / e^v) * (1 / (ln|u| + 1)) * dv.
Step 8: Simplify the integral further.
Since du = -dx, we have u = -x.
Therefore, |u| = |-x| = x, and ln|u| = ln|x|.
The integral looks as follows:
-∫ (1 / e^v) * (1 / (ln|x| + 1)) * dv.
Step 9: Finally, integrate the expression.
The integral of (1 / e^v) is -e^(-v).
So our final expression is:
-e^(-v) / (ln|x| + 1) + C,
where C is the integration constant.
Hence, the integral of x^-x is given by -e^(-u * ln|u|) / (ln|x| + 1) + C, with the limits going from 0 to -∞.