# How do you evaluate the integral #int x^1000e^-x# from 0 to #oo#?

you can do this very quickly by noting that it is the equivalent of

Or in terms of the gamma function:

To generate your own solution you could start as per the factorial function with this

Such that

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so we have

the recurrence equation reads

developping

or

so

or

Finally

so

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To evaluate the integral ∫x^1000e^(-x) from 0 to ∞, you can use integration by parts. Let u = x^1000 and dv = e^(-x) dx. Then, differentiate u to find du and integrate dv to find v. After that, apply the integration by parts formula:

∫u dv = uv - ∫v du

First, find du and v:

du = d(x^1000) = 1000x^999 dx v = ∫e^(-x) dx = -e^(-x)

Now, apply the integration by parts formula:

∫x^1000e^(-x) dx = -x^1000e^(-x) - ∫(-e^(-x))(1000x^999) dx

To evaluate the second integral, we can recognize that it's of the same form as the original integral. So we'll apply integration by parts again:

Let u = x^999 and dv = e^(-x) dx. Then, differentiate u to find du and integrate dv to find v.

du = d(x^999) = 999x^998 dx v = ∫e^(-x) dx = -e^(-x)

Now, apply the integration by parts formula again:

∫x^1000e^(-x) dx = -x^1000e^(-x) - ((-x^999)e^(-x) - ∫(-e^(-x))(999x^998) dx)

Simplify the expression:

= -x^1000e^(-x) + x^999e^(-x) - 999∫x^998e^(-x) dx

Now, we have another integral of the same form. We repeat the process until we get a term that's easy to evaluate. After several iterations, you'll eventually arrive at an expression that involves e^(-x) multiplied by a polynomial term, which tends to 0 as x approaches infinity. Thus, the integral converges to 0 as the upper limit approaches infinity.

Therefore, the value of the integral ∫x^1000e^(-x) from 0 to ∞ is 0.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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