Integration of natural logs.

[scguy]

From example 2:

I can kind of understand the method used but i just don't know why that once the product rule has been used there is no way to just integrate the 2 terms. I know that it is the same term that we started with and that it has just been left in its integrand notation to keep things simple, i think that is correct. I have written on the RHS what i think the method is i would just like to know if i am on the right track, which if i am not is not really surprising to me

[Johnny5]

I am not sure I understand your question.

 

I think you are just asking how to integrate x times e raised to the 2x power.

 

If that is all, there are several ways you can do this, I am not sure if it matters to you how you solve the problem.

 

The first thing that should jump out at you, is that you can solve it using integration by parts. Another method, would be to convert the integrand to a series, then integrate term by term, and then convert the final answer back into elementary functions.

 

I can do it both ways for you.

 

[math] \int x e^{2x} dx [/math]

 

Here is the integration by parts formula:

 

[math] \int u dv = uv - \int v du [/math]

 

In your specific problem, let u=x, and let v = 1/2 e^(2x)

 

From which it follows that:

 

du=dx and dv = e^(2x)dx

 

From which it follows that:

 

udv = x e^(2x)dx

 

From which it follows that:

 

[math] \int u dv = \int x e^{2x} dx [/math]

 

So by the transitive property of equality:

 

[math] \int x e^{2x} dx = uv - \int v du [/math]

 

From which it follows that:

 

[math] \int x e^{2x} dx = \frac{xe^{2x}}{2} - \frac{1}{2} \int e^{2x} dx [/math]

 

And the integral of e^(2x)dx is just 1/2 e^(2x).

 

Therefore:

 

[math] \int x e^{2x} dx = \frac{xe^{2x}}{2} - \frac{1}{4} e^{2x} + C[/math]

 

Where C is just an arbitrary constant of integration, and can be absorbed into the other constant.

 

 

And I promised you another method.

 

This method uses the fact that:

 

[math] e^x = \sum_{n=0}^{n=\infty} \frac{x^n}{n!} [/math]

 

For e^(2x), we have:

 

[math] e^{2x} = \sum_{n=0}^{n=\infty} \frac{(2x)^n}{n!} [/math]

 

Multiply both sides of the previous equation by x, to obtain:

 

[math] xe^{2x} = x\sum_{n=0}^{n=\infty} \frac{(2x)^n}{n!} [/math]

 

Summation is over n, and n is not dependent upon x, and therefore can pass through the summation sign. Therefore:

 

[math] xe^{2x} = \sum_{n=0}^{n=\infty} x \frac{(2x)^n}{n!} [/math]

 

Multiply both sides of the equation above by dx, to obtain:

 

[math] xe^{2x} dx = \sum_{n=0}^{n=\infty} x \frac{(2x)^n}{n!} dx [/math]

 

Now, integrate both sides of the previous equation, to obtain:

 

[math] \int xe^{2x} dx = \int \sum_{n=0}^{n=\infty} x \frac{(2x)^n}{n!} dx [/math]

 

Now, you can integrate term by term. The RHS is a series, so you have something like this:

 

integral of [a+b+c+d+e...]

 

and an integral of a sum, is the sum of the integrals, in other words:

 

[math] \int (a+b+c) = \int a + \int b + \int c [/math]

 

in words"integral of a sum is the sum of the integrals." This can be proven, but that is a different excercise.

 

back to the problem...

 

So we can pass the integral sign through the summation sign to obtain:

 

[math] \int xe^{2x} dx = \sum_{n=0}^{n=\infty} \int x \frac{(2x)^n}{n!} dx [/math]

 

Now, you see the (2x)^n quantity, and if the exterior was also (2x) then you could get (2x)^(n+1), so multiply by 2/2 to get:

 

[math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \int 2x \frac{(2x)^n}{n!} dx [/math]

 

Therefore:

 

[math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \int \frac{(2x)^{n+1} }{n!} dx [/math]

 

Therefore:

 

[math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \frac{1}{n!} \int (2x)^{n+1} dx [/math]

 

And the integral that must now be performed is trivial.

 

For example suppose that n=3, then the integral would be:

 

[math] \int (2x)^4 dx [/math]

 

Let U=2x, therefore dU=2dx, therefore 1/2dU=dx, now substitute, to obtain:

 

[math] \frac{1}{2} \int U^4 dU [/math]

 

Which is just 1/2( U^5/5 )

 

So we can make that substitution for any n, therefore:

 

Let U=2x

 

[math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \frac{1}{n!} \int U^{n+1} \frac{1}{2} dU[/math]

 

Therefore:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{1}{n!} \int U^{n+1} dU[/math]

 

Therefore:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{1}{n!} [ \frac{U^{n+2}}{n+2} + C]

[/math]

 

Replacing U by 2x, we obtain:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{1}{n!} [ \frac{(2x)^{n+2}}{n+2} + C]

[/math]

 

Whence it follows that:

 

 

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(2x)^{n+2}}{(n+2) n!} + C \sum_{n=0}^{n=\infty} \frac{1}{n!}} [/math]

 

Where I let the arbitrary constant c absorb the constant 1/4. Now, the integral is totally over. We can leave the answer as is, or we can attempt to write it in terms of elementary functions.

 

Although integration by parts was quicker, the method I've just shown is more powerful.

 

Let me attempt to write the infinite sums in terms of elementary functions.

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(2x)^{n+2}}{(n+2) n!} + Ce} [/math]

 

Where I have made use of the fact that:

 

[math] e = \sum_{n=0}^{n=\infty} \frac{1}{n!} = 2.71828...[/math]

 

Which is just a pure number, and can be absorbed in to the arbitrary constant of integration. So...

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(2x)^{n+2}}{(n+2) n!} + C} [/math]

 

Now, multiply both the numerator and denominator by (n+1), like so:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(n+1)(2x)^{n+2}}{(n+2)(n+1) n!} + C} [/math]

 

 

and (n+2)(n+1)n! is just n plus two factorial, that is:

 

(n+2)(n+1)n! = (n+2)!

 

Therefore:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(n+1)(2x)^{n+2}}{(n+2)!} + C} [/math]

 

From which it follows that:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(n-1)(2x)^n}{n!} + C} [/math]

 

From which it follows that:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{n(2x)^n}{n!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C}

 

 

[/math]

 

From which it follows that:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{(n-1)!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C}

 

 

[/math]

 

Where I simply converted n! into n(n-1)!, and then cancelled out n, from the numerator and denominator.

 

It now follows that:

 

[math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=1}^{n=\infty} \frac{(2x)^{n+1}}{n!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C}

 

 

[/math]

 

It follows that:

 

[math] \int xe^{2x} dx = \frac{1}{4} 2x \sum_{n=1}^{n=\infty} \frac{(2x)^n}{n!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C}

 

[/math]

 

We can now make use of the fact that:

 

[math] e^{2x} = \sum_{n=0}^{n=\infty} \frac{(2x)^n}{n!} = 1+2x+ \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} [/math]

 

And the previous fact leads to:

 

[math] e^{2x} -1 - 2x = \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} [/math]

 

Which we can now use above, to obtain:

 

[math] \int xe^{2x} dx = \frac{1}{4} 2x \sum_{n=1}^{n=\infty} \frac{(2x)^n}{n!} - \frac{1}{4} (e^{2x} -1 - 2x) + C} [/math]

 

And since:

 

[math] e^{2x} -1 = \sum_{n=1}^{n=\infty} \frac{(2x)^n}{n!} [/math]

 

It follows that:

 

[math] \int xe^{2x} dx = \frac{1}{4} 2x (e^{2x} -1) - \frac{1}{4} (e^{2x} -1 - 2x) + C} [/math]

 

From which it follows that:

 

[math] \int xe^{2x} dx = \frac{x}{2} e^{2x} - \frac{x}{2} -\frac{e^{2x}}{4} +\frac{1}{4} +\frac{x}{2} + C} [/math]

 

From which it now follows that:

 

[math] \int xe^{2x} dx = \frac{x}{2} e^{2x} -\frac{e^{2x}}{4} + C} [/math]

 

Which is the exact same answer we got using integration by parts.

 

QED

 

Regards

 

PS: By the way, the series solution was more for me to run through some things for myself, I recommend integration by parts for your original problem.

[scguy]

Got it, i know it is not really complicated or anything but i just wanted to check i was on the right track with the concept part.

U must be an expert using latex I will have to learn myself.

[Johnny5]

Got it' date=' i know it is not really complicated or anything but i just wanted to check i was on the right track with the concept part.

U must be an expert using latex I will have to learn myself.[/quote']

 

Here is a derivation of IBP formula.

 

Let u denote an arbitary function of x, thus u=u(x).

Let v denote an arbitrary function of x, thus v=v(x).

 

Now, consider their product:

 

uv

 

Differentiate uv with respect to x, using the product rule.

 

[math] \frac{d(uv)}{dx} = u\frac{dv}{dx} +v\frac{du}{dx} [/math]

 

Now, multiply both sides of the equation above by dx, to obtain:

 

[math] d(uv) = udv +vdu [/math]

 

Now, integrate both sides of the equation above to obtain:

 

[math] \int d(uv) = \int udv + \int vdu [/math]

 

Now, recognize that the LHS is an exact differential... "integral of d(anything)=anything +c

 

Thus:

 

[math] uv + C = \int udv + \int vdu [/math]

 

Ignore the arbitrary constant of integration, in the case of an indefinite integral, it can be absored at the end into just one, so it is redundant here. Thus:

 

[math] uv = \int udv + \int vdu [/math]

 

Thus:

 

[math] uv - \int vdu = \int udv [/math]

 

And since 'equality' is symmetric...

 

 

[math] \int udv = uv - \int vdu [/math]

 

Which was to be shown.

 

Regards