How do you evaluate the integral #int x^5dx# from #-oo# to #oo#?

Question

Scarlett Allison · Accepted Answer

To attempt to evaluate #int_-oo^oo f(x) dx# we choose a number #c# and evaluate, separately, #int_-oo^c f(x) dx# and #int_c^oo f(x) dx#

In order for #int_-oo^oo f(x) dx# to converge, both of the integrals on the half-lines must converge.

We'll use #c=0# (because it's easy to work with).

#int_-oo^0 x^5 dx = lim_(ararr-oo) int_a^0 x^5 dx#

# = lim_(ararr-oo)# #{: x^6/6]_a^0#

# = lim_(ararr-oo) a^6/6#

This limit does not exist, so the integral on the half-line diverges.

Therefore, the integral on the real line diverges.

Note

Since #x^5# is an odd function it is tempting to reason as follows.

For every positive number #a#, we have

#int_-a^a x^5 dx = 0#

So we must also have the integral from #-oo# to #oo# is #0#.

This reasoning FAILS because we define #int_-oo^oo f(x) dx# as above.

Adam Adkins · Answer

undefined To evaluate the integral ∫x^5 dx from -∞ to ∞, you can use the property of odd functions and the symmetry of the integrand. Since x^5 is an odd function (f(-x) = -f(x)), the integral from -∞ to 0 will cancel out with the integral from 0 to ∞, leaving only one part to evaluate. Thus, you can evaluate the integral from 0 to ∞ and then double the result.

∫x^5 dx from 0 to ∞ can be calculated using the power rule for integration. The integral of x^n dx is (1/(n+1))x^(n+1), so the integral of x^5 dx is (1/6)x^6.

Evaluate (1/6)x^6 from 0 to ∞:

∫(1/6)x^6 dx from 0 to ∞ = [(1/6)(∞^6)] - [(1/6)(0^6)]
= (1/6)(∞^6) - 0
= ∞

Since the integral from -∞ to ∞ is the sum of the integral from -∞ to 0 and the integral from 0 to ∞, and each part is the same, the total integral is twice the integral from 0 to ∞, which is ∞. Therefore, the integral of x^5 dx from -∞ to ∞ is ∞.