integration formulae

[PG1995]

Hi

Could you please help me with the query included in the attachment? Thank you.

Regards
PG

 

[steveB]

The difference between the top formulas and the corresponding bottom formulas is that the bottom formulas are more general and include a scale factor "a" in the function. So if a=1, you get the top formulas, and if "a" not equal to 1 then you need the more general formulas.

 

[MrAl]

Integral e^u du=e^u

if u=a*x then:
Integral e^u du = Integral e^(a*x) du, where u=a*x, so du/dx=a so du=a*dx so:
Integral e^u du = Integral e^(a*x)*a dx= e^(a*x)=e^u

So when we sub du for dx we have to use a*dx so the 'a' cancles out so:

Integral e^u du = Integral e^ax*a dx = e^ax = e^u

 

[PG1995]

Thank you, Steve, MrAl.

The difference between the top formulas and the corresponding bottom formulas is that the bottom formulas are more general and include a scale factor "a" in the function. So if a=1, you get the top formulas, and if "a" not equal to 1 then you need the more general formulas.

What do you mean by "scale factor"? Do you mean to say that the function f(x) being multiplied by "a" as in e^{a*f(x)}.

I was playing around with my calculator where I entered this expression just randomly. Is there any simple way to evaluate it? It seems it's difficult to integrate because my calculator only simplifies it to: [latex]$7.39\int e^{2x^{2}}dx$[/latex], and a software package also fails to simplify it. Please help me with it. Thank you.

Regards
PG

 

[steveB]

Thank you, Steve, MrAl.



What do you mean by "scale factor"? Do you mean to say that the function f(x) being multiplied by "a" as in e^{a*f(x)}.



I was playing around with my calculator where I entered this expression just randomly. Is there any simple way to evaluate it? It seems it's difficult to integrate because my calculator only simplifies it to: [latex]$7.39\int e^{2x^{2}}dx$[/latex], and a software package also fails to simplify it. Please help me with it. Thank you.

Regards
PG

A scale factor is just a constant that multiplies a variable. In this case the letter "a" is clearly shown in your formulas 14.339 and 14.509. So, there should be no confusion about what I mean by saying that these formulas include a scale factor "a". If you set a=1, the you get formulas 14.9 and 14.11. It's as simple as that.

Now, the integral you are trying to evaluate is not of the form of any of these formulas. The reason why you are having trouble is that there is no simple formula for the integral you are trying to evaluate. Your attempt to try a substitution that might put you in a form to use 14.339 or 14.509 did not work. Actually, you made a mistake to put the 1/(4x) outside the integral. It must stay inside the integral and then you write 1/(4x) in terms of "u". You can be sure that whatever you end up with will just be another integral that does not have a nice simple formula.

 

[MrAl]

Hi,


When you use a 'u' substitution you have to have the du inside the integral or else can make it that way with a constant. For example:
x*e^(x^2)

can be integrated using 'u' substitution because when you take the first derivative of x^2 you get 2*x, and 'x' is inside the integral too, so you can do:
du/dx=2*x
du=2*x*dx
and since we had
x*e^u
put this inside the integral:
e^u du
but since we had a '2' in the du we have to take the constant 1/2 outside the integral, so we have:
(1/2)*Integral (e^u du) = (1/2)*e^u + C = e^(x^2)/2 + C

But yes, that's not the integral you are trying to do, as the integral of e^(x^2) even by itself is not a simple integration as Steve says.
I think this is one of those where you need to use a table, or possibly you can only do this one numerically. I think you can break it down into another function like a constant times Integral[0 to w] e^(-t^2) dt where w is itself a constant times x and possibly not real, but i think you have to integrate that numerically anyway (note the change in exponent sign). There are approximations for that new integral that go by the name of "erf(x)", and so you could use one of those but it will be an approximation anyway.

There are some integrals that you just can not do except by numerical integration, others are very difficult. That's why it is good to know some different methods of numerical integration too. Numerical integration is also good for checking results of integrations you do other ways.

 

[panic mode]

I was playing around with my calculator where I entered this expression just randomly. Is there any simple way to evaluate it?

does not get any simpler than letting wolfram alpha take a look at it. also you can click on "show steps" to check how exactly it arrived to solution.

https://www.wolframalpha.com/input/?i=integral+of+e^(2x)+dx

 

[MrAl]

Hi,

Yes good idea there.

I just did e^(2*x^2) dx and got:
(1/2)*sqrt(pi/2)*erfi(sqrt(2)*x)+C

You'd have to next look up erfi(x) to calculate that part of it, most likely it has to be done numerically, expansion, or known approximation.

 

[PG1995]

Thank you, Steve, MrAl, Panic Mode.

A scale factor is just a constant that multiplies a variable. In this case the letter "a" is clearly shown in your formulas 14.339 and 14.509. So, there should be no confusion about what I mean by saying that these formulas include a scale factor "a". If you set a=1, the you get formulas 14.9 and 14.11. It's as simple as that.

No, there was actually no confusion about what you wrote. I was looking at it in the context of the integral, ∫e^{2(x^2+1)}dx. I thought perhaps I can simply write: ∫e^{2(x^2+1)}dx = e^{2(x^2+1)}/2. Obviously I was wrong. Besides, the expression "(x^2+1)" isn't a variable, it's a function in itself.

@Panic Mode, MrAl:
It's really a wild animal and at this level I don't intend to domesticate it! I get this:



Regards
PG

 

[panic mode]

fair enough, even if erf and flavours are not complicated
https://en.wikipedia.org/wiki/Erfi

 

[steveB]

Obviously I was wrong. Besides, the expression "(x^2+1)" isn't a variable, it's a function in itself.

Well, variable or function can essentially be the same thing. What you attempted to do was actually ok, at the start. You tried to define u=2(x^2+1), right? So, the expression was a variable "u", and there is no problem with that. You then needed to carry out the substitution to see if you might arrive at a form you can solve. This is where you made an error, by factoring the 1/(4x) out of the integral. If you leave it inside the integral and then realize that x=sqrt(u-1), you would have a new integral to solve. It just so happens that this integral is also not solvable.

So, you were on the right track, considering you are not familiar enough with this integral to know it is not solvable. Many mathematicians tried to solve this in the past, and failed. So, I wouldn't say you were wrong conceptually, but rather you made a mistake along the way.

 

[PG1995]

Hi

Could you please help me with this query? Thanks a lot.

Regards
PG

 

[steveB]

Hi

Could you please help me with this **broken link removed**? Thanks a lot.

Regards
PG

As you said, they could have said it is a continuous function, but this is a weaker statement than saying it is a piecewise continuous function. A continuous function can not have any sudden jumps in it, but a piecewise continuous function can have any finite number of sudden jumps.

For example, a unit step function is a piecewise continuous function, but it is not a continuous function if you try to integrate through t=0.

So the integral of a unit step from -1 to +1 is a limit that will exist, and the answer is +1. Your weaker statement would have nothing to say about this integral. The stronger statement has a definite thing to say about this other integral and others like it.

 

[MrAl]

Hi,

Actually a piecewise continuous function can NOT have sudden jumps in it, it can only have sudden turns.
So that's why they wanted to say piecewise continuous, so the function requirement is less constrictive.

A piecewise function can have both sudden jumps and sudden turns.
So a piecewise function and a piecewise continuous function are two different animals.

By stating that the function is continuous means that it is differentiable everywhere, whereas stating the function is piecewise continuous means that it is differentiable only within the segments but the limits of the endpoints of each segment are still the same as with a continuous function.

So the function they are talking about is not continuous but it is piecewise continuous.

A simple example of a piecewise continuous function is
y= abs(x)
where at the origin the left sided limit is equal to the right sided limit even though the derivative doesnt exist.

A simple example of a piecewise (but not continuous) function is:
y=2 {x<0}
y=1 {x>=0}
where the right sided limit is 2 but the left sided limit is 1, so the function is piecewise but NOT continuous.

So in short you cant say that a function is continuous unless at all points the right sided limit is equal to the left sided limit.

So we have three major classes of functions:
1. Continuous: derivative everywhere and equal left and right limits everywhere.
2. Piecewise: derivative only within segments, and left and right limits may be unequal at end points.
3. Piecewise continuous: derivative only within segments, and left and right limits must be equal everywhere.


BTW, for the previous question this reference might be interesting:
https://en.wikipedia.org/wiki/Normal_distribution

 

[PG1995]

Thank you, Steve, MrAl.

Actually a piecewise continuous function can NOT have sudden jumps in it, it can only have sudden turns.
So that's why they wanted to say piecewise continuous, so the function requirement is less constrictive.

A piecewise function can have both sudden jumps and sudden turns.
So a piecewise function and a piecewise continuous function are two different animals.

The quoted section above is the gist of it. In my view, continuous functions are a subset of a piecewise continuous functions and that's the reason the book used the terminology it did.

Regards
PG

 

[MrAl]

Thank you, Steve, MrAl.



The quoted section above is the gist of it. In my view, continuous functions are a subset of a piecewise continuous functions and that's the reason the book used the terminology it did.

Regards
PG

Hi,


Well, the theory holds for continuous functions but since it also holds for piecewise continuous functions they knew they could include those too. If they said just continuous functions then they would have been leaving out many functions that also work in the theory. But they did not just say 'piecewise' without the 'continuous' which means they wanted to exclude functions that are solely piecewise because the theory doesnt hold for them.
Is that what you thought about too?

 

[PG1995]

Thanks, MrAl.

Is that what you thought about too?

Yes.

Regards
PG

 

[steveB]

I have to disagree with the above. A piecewise continuous function is usually defined to be that which is continuous within a finite number of regions, and then finite discontinuities are allowed at the boundaries between pieces.

I expect the reference cited by PG is using this definition, but I guess PG needs to verify by reading the reference.

https://www.electro-tech-online.com/custompdfs/2012/10/pg13.pdf

If you think about it, there is no issue integrating through a finite number of discontinuities. One simply breaks the full integral into a finite number of integrals over each continuous piece.

 

[MrAl]

I have to disagree with the above. A piecewise continuous function is usually defined to be that which is continuous within a finite number of regions, and then finite discontinuities are allowed at the boundaries between pieces.

I expect the reference cited by PG is using this definition, but I guess PG needs to verify by reading the reference.

https://www.electro-tech-online.com/custompdfs/2012/10/pg13.pdf

If you think about it, there is no issue integrating through a finite number of discontinuities. One simply breaks the full integral into a finite number of integrals over each continuous piece.

Hi Steve,

Well strictly speaking a jump is called a jump discontinuity, so that in itself should tell us something. And there is no way to define a curve with a jump to make it continuous which is not the case with a removable discontinuity. In fact, that's one thing about a jump discontinuity that is known, that it can not be removed, while some other types of discontinuities can be removed. So what is it that we are attempting to remove then if it's not a discontinuity?

As im sure you know, a curve that is defined over some range of x but has even one point removed is not continuous. That point has to be marked as an open circle rather than a point. A jump however can be labeled a few different ways:
1. Open on the left, a point on the right.
2. Open on the right, a point on the left.
3. Open on the left and on the right (conditional).
4. A point on the left and a point on the right perhaps.
5. Open on the left, open on the right, somewhere in between the two vertically there is a point.

In all of these the limits on the left do not equal the limits on the right, and also for some the value of the function does not equal the limit at that point.

We can even have any of 1 to 4 above but have a horizontal space between the two points.

We might be able to integrate some of these but not all of them, at least not without cheating in some way.

In real life we can often overlook the stricter view because those tiny places dont mean much, but in pure math it's defined very strictly.

As another finer point, if a function is not defined somewhere it may still be considered continuous (i think this is rare though so we should probably ignore this case as it would still be correct to say 'discontinuous'):
y=2, x<0
y=1, x>0

but not:
y=2, x<0
y=1, x>=0

 

[steveB]

So are you saying the definition I provided is not correct?

Ive seen the definition of "piecewise continuous" perhaps nearly 100 times in my life in lectures and textbooks. Always the definition allows for finite number of discontinuities with a very basic constraint about the limit of the values.

Of course, what matters is the definition in PGs book, so I would ask that he post that.