Integration by Parts: A Powerful Tool in the Theory of Distributions

[Whatupdoc]

$$\int 4x cos(2x)$$

using integration by parts...

u=4x
du= 4
dv=cos(2x)
v=cos(x)sin(x)

using the formula uv - $$\int$$v*du...

4x*cos(x)sin(x) - $$\int$$cos(x)sin(x)*4x

hmm i can't seem to finish this problem, can someone help? and am i doing it correctly so far?

 

[Tide]

Your expression for v is incorrect.

 

[arildno]

DO NOT USE THIS HORRIBLE MIXTURE OF INTEGRATION BY PARTS AND INTEGRATION BY SUBSTITUTION!
Use the following:
$$\int{uv'dx}=uv-\int{u'vdx}+C$$
Set:
$$u(x)=4x, v'(x)=\cos(2x)\to{v(x)}=\frac{1}{2}\sin(2x)$$

 

[Whatupdoc]

DO NOT USE THIS HORRIBLE MIXTURE OF INTEGRATION BY PARTS AND INTEGRATION BY SUBSTITUTION!
Use the following:
$$\int{uv'dx}=uv-\int{u'vdx}+C$$
Set:
$$u(x)=4x, v'(x)=\cos(2x)\to{v(x)}=\frac{1}{2}\sin(2x)$$

ok so i plugged the parts into the formula
$$\int{uv'dx}=uv-\int{u'vdx}+C$$

$$4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C$$

anti-deri. of sin is -cos that is why it's adding now...

4x*1/2*sin(2x) + 4x*1/4*cos(2x)
so... would that be the answer?

 

[Ethereal]

ok so i plugged the parts into the formula
$$\int{uv'dx}=uv-\int{u'vdx}+C$$

$$4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C$$

anti-deri. of sin is -cos that is why it's adding now...

4x*1/2*sin(2x) + 4x*1/4*cos(2x)
so... would that be the answer?

Yes, it's the answer, try differentiating it.

EDIT: Error.

 

[Whatupdoc]

the thing is, the answer is incorrect. i don't know what the correct answer is, but i know that isn't the answer. i submit my homework thru a online server and it checks your answer. what am i doing wrong?

 

[Ethereal]

the thing is, the answer is incorrect. i don't know what the correct answer is, but i know that isn't the answer. i submit my homework thru a online server and it checks your answer. what am i doing wrong?

My deepest apologies. I must have been blind! The correct answer should be $$2xsin2x + cos2x$$ (written in Tex for clarity)

 

[Whatupdoc]

thanks for the correct answer, but where did i go wrong?

 

[Ethereal]

4x*1/2*sin(2x) + 4x*1/4*cos(2x)

That x is not supposed to be there.

 

[Whatupdoc]

$$4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C$$

hmm i thought i was suppose to take the anti-derivative of 4 also... can you tell me why i don't need to take the anti-derivative of the "4" after the intergral?

 

[Math Is Hard]

here are my thoughts..

$$\int 4*1/2*sin(2x)dx$$
is equivalent to
$$\int 2 *sin(2x)dx$$
is equivalent to
$$2* \int sin(2x)dx$$

you can always pull out the constant factor

 

[HallsofIvy]

$$4x*1/2*sin(2x) - \int 4*1/2*sin(2x)*dx + C$$

hmm i thought i was suppose to take the anti-derivative of 4 also... can you tell me why i don't need to take the anti-derivative of the "4" after the intergral?

What do you mean by "take the anti-derivative of 4 also"? Did you also think you should take the anti-derivative of the 1/2?
$\int{f(x)g(x)dx}$ is NOT $\int{f(x)dx}\int{g(x)dx}$.

In particular, if C is any constant (4, 1/2, whatever),
[tex]\int{Cf(x)dx}= C\int{f(x)dx}[/itex].<br /> <br /> $$\int{4(1/2)sin(2x)dx}= \int{2sin(2x)dx}= 2\int{sin(2x)dx}= -cos(2x)+ C$$.[/tex]

 

[Whatupdoc]

What do you mean by "take the anti-derivative of 4 also"? Did you also think you should take the anti-derivative of the 1/2?
$\int{f(x)g(x)dx}$ is NOT $\int{f(x)dx}\int{g(x)dx}$.

In particular, if C is any constant (4, 1/2, whatever),
[tex]\int{Cf(x)dx}= C\int{f(x)dx}[/itex].<br /> <br /> $$\int{4(1/2)sin(2x)dx}= \int{2sin(2x)dx}= 2\int{sin(2x)dx}= -cos(2x)+ C$$.[/tex]

[tex] <br /> <br /> that was very very very helpful, thank you sooo much. i always thought that $\int{f(x)g(x)dx}$ is equal to $\int{f(x)dx}\int{g(x)dx}$ . <br /> <br /> ok one more thing, so what is $\int{f(x)g(x)dx}$ then? would i just take the anti-derv. of f(x)?[/tex]

 

[Tide]

No. Generally, you would have to find the antiderivative of $f(x)g(x)$. Knowing the antiderivative of one of them would permit you to use integration by parts.

 

[Ethereal]

This is a little off-topic, but given a function, how do we know if it's anti-derivative exists?

 

[Muzza]

I think all continuous functions have anti-derivatives, because given a real continuous function f defined on say [a, b], if we define A by

$$A(x) = \int^{x}_{a} f(t) dt$$

then A'(x) = f(x) (for all sensible values of x) (this is usually part of the proof of the fundamental theorem of (integral) calculus). So an anti-derivative exists (actually writing such a function in a closed form is a totally different matter though ;)).

Someone correct me if I'm wrong...

 

[HallsofIvy]

This is a little off-topic, but given a function, how do we know if it's anti-derivative exists?

Every continuous function (in fact, every bounded function whose points of discontinuity form a set of measure 0) has an anti-derivative. Whether or not that anti-derivative is an "elementary" function depends on your definition of "elementary" function!

that was very very very helpful, thank you sooo much. i always thought that $\int{f(x)g(x)dx}$ is equal to $\int{f(x)g(x)dx}$.

ok one more thing, so what is $\int{f(x)g(x)dx}$ then? would i just take the anti-derv. of f(x)?

No, of course, not! You can't just ignore g! The way you integrate a product IS "integration by parts".


Remember the differentiation laws? IF (fg)'= f' g' THEN we could integrate both sides and get $\int{f(x)g(x)dx}=\int{f(x)g(x)dx}$.

But that's NOT true. What is true is the "product rule": (fg)'= f'g+ fg'.
That's the same as fg'= (fg)'- f'g. Integrating both sides,
$\intf(x)g'(x)dx= fg- \int g f' dx$.

Writing u= f, dv= g'dx, that is exactly $\int udv= uv- \int vdu$.

 

[Whatupdoc]

No, of course, not! You can't just ignore g! The way you integrate a product IS "integration by parts".


Remember the differentiation laws? IF (fg)'= f' g' THEN we could integrate both sides and get $\int{f(x)g(x)dx}=\int{f(x)g(x)dx}$.

But that's NOT true. What is true is the "product rule": (fg)'= f'g+ fg'.
That's the same as fg'= (fg)'- f'g. Integrating both sides,
$\intf(x)g'(x)dx= fg- \int g f' dx$.

Writing u= f, dv= g'dx, that is exactly $\int udv= uv- \int vdu$.

i read your post many many times, and i kinda get what your saying, but i am also confused about some parts. I am way better with examples, can you help me step by step and solve this problem:

$$\int 4x cos(2x)$$ <-- this problem is the one i asked for from my first post, i already got the answer, but i don't know how Ethereal got the answer.

let me show you how i tried to solve the problem.

using $$\int{uv'dx}=uv-\int{u'vdx}+C$$
u(x) = 4x
du(x) = 4dx
dv(x) = cos(2x)
v = 1/2sin(2x)

plugged in what i have...

$$4x*1/2sin(2x) \int 4*1/2sin(2x)$$

ok now i know that i need to solve $$\int 4*1/2sin(2x)$$
ok at this point, i am really stuck and confused. can someone help me step by step and explain to me what i really need to do? i know it's a lot to ask for, but i really want to learn it.

 

[Math Is Hard]

I think there's still some problems in your setup. The formula for int. by parts is:
$\int udv= uv- \int vdu$

Let u = 4x
dv = cos(2x)dx
du = 4 dx
v = 1/2 sin(2x)

$\int 4x cos(2x)dx = 4x (1/2 sin (2x)) - \int (1/2 sin (2x)) 4 dx$

which becomes:

$\int 4x cos(2x)dx = 2x sin (2x) - \int 2 sin (2x) dx$

which becomes

$\int 4x cos(2x)dx = 2x sin (2x) - 2\int sin (2x) dx$

the last integral is pretty easy to solve with a substitution for 2x. Does that help at all?

 

[Math Is Hard]

If you're stuck on the last part, show how you would integrate this:

$\int sin(2x)dx$

that should pinpoint where the problem is.

 

[Whatupdoc]

thanks for the help, everything makes sense now besides one thing.that is extactly where I am having the problem at, would i use the chain rule backwards or something to integrate $\int sin(2x)dx$ ?

and do you see how $\int (1/2 sin (2x)) 4 dx$ became $\int 2 sin (2x) dx$. if 4 was let's say 4x, would i have to do a second integration for the 2nd part of the problem?

 

[Ethereal]

Every continuous function (in fact, every bounded function whose points of discontinuity form a set of measure 0) has an anti-derivative. Whether or not that anti-derivative is an "elementary" function depends on your definition of "elementary" function!

Is there a proof of this somewhere? And is there a standard definition which exists for "elementary"?

 

[Math Is Hard]

you can use substitution on

$\int sin(2x)dx$

let u = 2x
du = 2 dx
1/2 du = dx

now you can rewrite the integral like so:

$\int sin u (1/2 du)$

or

$1/2 \int sin u du$

which is 1/2 * (-cos u) + c

and subbing back is

-1/2 cos(2x) + c

(apologies for my cruddy latex)

 

[Whatupdoc]

ah perfect, i totally understand. thank you

 

[HallsofIvy]

Is there a proof of this somewhere? And is there a standard definition which exists for "elementary"?

Proofs that every continuous function or every bounded function with a finite number of discontinuities can be found in any good calculus book. The proof that every bounded function whose discontinuities form a set of measure is (Riemann) integrable depends upon "measure" theory, normally an advanced topic in analysis, in order to define "set of measure zero"!

In terms of the Lebesque integral, there exist functions discontinuous everywhere that are integrable.

 

[mathwonk]

halls of ivy's interesting remarks, may be enlarged on as follows.

first of all it is a basic fact of elementary integration theory, proved say in spivak's calculus on manifolds, that every bounded function f which has only a measure zero set of discontinutities, is riemann integrable.

then we can take the indefinite integral of any such function as an "antiderivative". i.e. the function G(x) = integral of f from a to x, is in some sense an "antiderivative" of f. I.e. the usual elementary proof of the fundamental theorem of calculus surely found in every calculus book, (except maybe sylvanus p thompson?), proves that G is differentiable at least at every point x where f was continuous, and at such points that G'(x) = f(x).

Moreover G is continuous everywhere.

However, it does not follow that such an "antiderivative" can be used to evaluate the integral of f. I.e. there may exist other continuous functions H which are also differentiable everywhere that f is continuous, and such that H'(x) = f(x) at such points, and yet H and G may not differ by a constant!

So the usual mean value theorem fails here for these more general functions.

So we would like to rule out allowing the latter function H be an "antiderivative" of f.

E.g. consider the famous "middle third" construction, in which we define H to equal 1/2 on the open middle third of the unit interval, to equal 1/4 on the open middle third of the interval from 0 to 1/3, and to be 3/4 on the open middle third of the interval from 2/3 to 1. Continue this construction forever, and obtain a function H which is continuous on all of [0,1], has H(0) = 0, H(1) = 1, but H has derivative zero on the union of all the middle third sets, which together have length one, by the geometric series formula.

Thus this H is an antiderivative in our naive sense above for the function f which equals zero on the union of the middle third sets and equals 1 elsewhere (elsewhere being a closed set of measure zero). However the integral of this function f is zero, while using H to compute H(1)-H(0) gives us 1.

Thus we need a morev restrictive definition of "antiderivative" if we want the concept to be useful in integrating arbitrary riemann integrabele functions. the concept needed is a strengthening of uniform continuity called "absolute continuity". i think it says more or less that given any length e, we can find a length d such that on any finite disjoint union of subintervals of total length d, the total change in the function is less than e.

Then if we define an antiderivative of a riemann integrable function f to be an absolutely continuous function G for which G'(x) = f(x) wherever f is continuous, then it does follow that any two such functions differ by a constant, and thus that the integral of f from a to b is G(b)-G(a), for anyone of them.


In trying to offer new stuff to my calculus clas this eyar I was venturing into these waters, and actually made a false conjecture about the more naive antiderivatives above, until educated by an analyst colleague. so to my great delight, i finally found out what that concept "absolute continutiy" was good for, that I encountered so long ago in measure theory!

by the way a set has measure zero if it can be covered by a sequence of intervals of arbitrarily small total length. for instance any sequence of points ahs measure zero, since given a length e, we can take a sequence of intervals of lengths e/2, e/4, e/8,... and center one interval on each point thus covering them all.

the complement of the middle thiord set disacussed above is actually more numerous than this, but has measure zero because the complement in [0,1] has length 1. i have proved elsewhere on this site that this set is uncountable.

 

[mathwonk]

integration by parts is extremely interestijng for another reason, it let's you differentiate functions which are not differentiable at all in any usual sense.

i.e. just as a vector in R^n is determined by its dot product with every other vector, a wide variety of functions are essentially determined by their analogous "dot products", i.e. by all the integrals of their products with all other smooth functions.

we can even restrict these other smooth functions to vanish at infinity. now imagine that there is a function we want to differentiate, but it is not differentiable, but it is at least integrable. Then by the remark abbove we would know what its "derivative" was if we knew the integral of that derivative against every smooth function vanishing at infinity.

But then by the integration by parts formula, the integral of our unknown "derivative" against that smooth function, should equal the (negative of) the integral of the original integrable function multiplied by the actual derivative of the smooth function!

i.e. integration by parts let's us move the derivative operation over from the function where it is not understood, to the one where it is known.

so although we cannot differentiate our function directly we can define the dot product with every smooth function (vanishing at infinity) of its so called "derivative".

so even though we do not know what the derivative is, we do know how it behaves under dot product with other smooth functions, and that is enough for many purposes.

gussied up, this is called the theory of "distributions", and is extremely powerful in proving existence of solutions of differential equations. see frank warner's book. e.g.