- 06 Sep '16 12:47 / 6 editsI am trying to find this integral and I tried using integration by parts but cannot seem to make it work for me. Then I tried wolfalpha and that gave me nothing. "{...}" indicates subscript.

S, F ∈ ℕ{1}

x ∈ ]0, 1[ , x ∈ ℝ{>0}

∫[0, x] X^S (1 – X)^F dX

So S and F are natural numbers excluding 0.

x and X are real numbers between 0 and 1 excluding 0 and 1.

All numbers positive. - 06 Sep '16 16:04

This is the incomplete Beta Function and there's some stuff about it on the Wikipedia page. For integer powers it looks as if you should be able to integrate by parts repeatedly:*Originally posted by humy***I am trying to find this integral and I tried using integration by parts but cannot seem to make it work for me. Then I tried wolfalpha and that gave me nothing. "{...}" indicates subscript.**

S, F ∈ ℕ{1}

x ∈ ]0, 1[ , x ∈ ℝ{>0}

∫[0, x] X^S (1 – X)^F dX

So S and F are natural numbers excluding 0.

x and X are real numbers between 0 and 1 excluding 0 and 1.

All numbers positive.

Let S = p - 1 and F = q - 1

I = ∫[0, x] t^(p - 1) (1 – t)^(q - 1) dt

= [1/p t^p (1 - t)^(q - 1)][x, 0] + (q - 1)/p ∫[0, x] t^p (1 – t)^(q - 2) dt

= (1/p)x^p(1 - x)^(q - 1) + (q - 1)/(p(p + 1)) x^(p + 1)(1 - x)^(q - 2) + ((q - 1)(q - 2))/(p(p+1)) ∫[0, x] t^(p + 1)(1 – t)^(q - 3) dt

= (1/p)x^p(1 - x)^(q - 1) + (q - 1)/(p(p + 1)) x^(p + 1)(1 - x)^(q - 2) + ... + ((q - 1)(q - 2) ... 2.1)/(p(p + 1) .... (p + q - 1))∫[0, x] t^(p + q - 1) dt

= (1/p)x^p(1 - x)^(q - 1) + (q - 1)/p x^(p + 1)(1 - x)^(q - 2) + ... + ((q - 1)(q - 2) ... 2.1)/(p(p + 1)(p + 2) .... (p + q - 1)(p + q))x^(p + q)

I typed this in without being very careful, so I'd check this derivation. But as long as one of p or q is a positive integer you can repeatedly integrate by parts until the integral is trivial. You can also do the integral using the binomial expansion of (1 - t)^(q - 1), although that's probably how you got to the integral in the first place...

https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function - 06 Sep '16 19:10 / 6 edits

I am frustrated here.*Originally posted by DeepThought***This is the incomplete Beta Function and there's some stuff about it on the Wikipedia page. For integer powers it looks as if you should be able to integrate by parts repeatedly:**

Let S = p - 1 and F = q - 1

I = ∫[0, x] t^(p - 1) (1 – t)^(q - 1) dt

= [1/p t^p (1 - t)^(q - 1)][x, 0] + (q - 1)/p ∫[0, x] t^p (1 – t)^(q - 2) dt

= (1/p)x^p(1 - x)^(q ...[text shortened]... al in the first place...

https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function

I know from experience you always know exactly what you are talking about but I don't understand your notation you are using here. For example, what does;

" [1/p t^p (1 - t)^(q - 1)][x, 0] "

mean? I am confused by the application of those square brackets.

Where did that expression come from?

How does it relate to integration by parts?

And I know the integration by parts formula of;

∫ uv dx = u ∫ vu dx - ∫ ( u' ∫ v dx ) dx

which is the one I tried to use first. Is that the one you are using in disguise?

Is there a reason why I cannot solve it with that formula?

I already tried and failed but would like to know if I failed because for some reason it cannot be done with that formula or if I am doing something wrong with that formula. If doing something wrong with that formula, I will go all though it again and try again harder.

I have looked at you link and others on the incomplete Beta Function but, and I am probably being stupid here, failed to see anything I know how to use to help actually*solve*the integral. - 06 Sep '16 20:11

The square bracket notation is meant to be read as:*Originally posted by humy***I am frustrated here.**

I know from experience you always know exactly what you are talking about but I don't understand your notation you are using here. For example, what does;

" [1/p t^p (1 - t)^(q - 1)][x, 0] "

mean? I am confused by the application of those square brackets.

Where did that expression come from?

How does it relate to integration by par ...[text shortened]... tupid here, failed to see anything I know how to use to help actually*solve*the integral.

[f(x)][a, b] = f(b) - f(a). The usual notation for what I was trying to do puts a and b as sub and superscripts after the first bracket, I was just borrowing your notation from the integral.

I don't recognize your formula for integration by parts. First an identity:

∫[a, b] df/dx dx = ∫[f = f(a), f = f(b)] df = f(b) - f(a) = [f(x)][a, b]

which is the Fundamental Theorem of Calculus. From the product rule for calculus we have:

d(uv)/dx = udv/dx + (du/dx)v = uv' + u'v

So, integrating we have:

∫[a, b] dx (d(uv)/dx) = ∫[a, b] dx (uv' + u'v)

We can use our identity on the left hand side to get:

∫[a, b] dx(d(uv)/dx) = [u(x)v(x)][a, b] = u(b)v(b) - u(a)v(a)

Putting this back into our integral expression and rearranging we get the usual form:

∫[a, b] dx u(x)(dv(x)/dx) = [u(x)v(x)][a, b] - ∫[a, b] dx (du(x)/dx)v(x)

or using the prime notation for derivatives:

∫[a, b] dx u(x)v'(x)= u(b)v(b) - u(a)v(a) - ∫[a, b] dx u'(x)v(x)

The application of this formula is what is normally meant by "Integration by Parts". - 07 Sep '16 07:00my formula for integration by parts is the first one listed under the subtitle;

"Application to find antiderivatives"

at;

https://en.wikipedia.org/wiki/Integration_by_parts

But I probably should give the whole subject area a major revision which could take me what will seem like an eternity. - 07 Sep '16 16:06

I see, whoever wrote that page uses dreadful notation. In your formula replace v with v' and it should turn into my form:*Originally posted by humy***my formula for integration by parts is the first one listed under the subtitle;**

"Application to find antiderivatives"

at;

https://en.wikipedia.org/wiki/Integration_by_parts

But I probably should give the whole subject area a major revision which could take me what will seem like an eternity.

v <----> v'

∫ uv dx = u ∫ v dx - ∫ ( u' ∫ v dx ) dx <---> ∫ uv' dx = u ∫ v' dx - ∫ ( u' ∫ v' dx ) dx = uv - ∫ u' v dx

They have v(x) doing two different jobs in that article and it's unhelpful. The notation I used (apart from coping with subscripts and superscripts using [a, b]) is standard and the notation they use in that section is non-standard. See the section on the Talk page headed "memnotechnics" - that is the right way to formulate integration by parts - and all you need to know - most of the rest of the page is a waste of time, definitely ignore their stupid mnemonics for choosing which function is u(x) and which v'(x), all professional mathematicians use their experience and ingenuity and not a daft mnemonic. You just need to practise a little, you already know what is easier to integrate and what to differentiate. In their notation ∫ (...) dx is sometimes being used as an antiderivative operator (and is an abuse of notation) and sometimes as a definite integral and it is confusing. The whole thing is a fairly horrible piece of writing so if you do feel like doing some revision then use a different site or buy an elementary (i.e. A-level) calculus book, which will hopefully have worked problems in. - 07 Sep '16 20:49

So one thing you are saying is that I should use;*Originally posted by DeepThought***I see, whoever wrote that page uses dreadful notation. In your formula replace v with v' and it should turn into my form:**

v <----> v'

∫ uv dx = u ∫ v dx - ∫ ( u' ∫ v dx ) dx <---> ∫ uv' dx = u ∫ v' dx - ∫ ( u' ∫ v' dx ) dx = uv - ∫ u' v dx

They have v(x) doing two different jobs in that article and it's unhelpful. The notation I used (apar ...[text shortened]... or buy an elementary (i.e. A-level) calculus book, which will hopefully have worked problems in.

∫ uv' dx = uv - ∫ u' v dx

and NOT

∫ uv dx = u ∫ v dx - ∫ ( u' ∫ v dx ) dx

I have been looking at some examples of using ∫ uv' dx = uv - ∫ u' v dx

at http://mathworld.wolfram.com/IntegrationbyParts.html

and it looks straightforward enough to me. - 08 Sep '16 00:45

It's not compulsory, other than the notational clumsiness of using the same symbol for anti-differentiation as for (definite) integration it's perfectly correct, but the first form is the standard one basically because the connection with the fundamental theorem of calculus is more evident.*Originally posted by humy***So one thing you are saying is that I should use;**

∫ uv' dx = uv - ∫ u' v dx

and NOT

∫ uv dx = u ∫ v dx - ∫ ( u' ∫ v dx ) dx

I have been looking at some examples of using ∫ uv' dx = uv - ∫ u' v dx

at http://mathworld.wolfram.com/IntegrationbyParts.html

and it looks straightforward enough to me. - 08 Sep '16 16:26 / 7 editsARR I have finally GOT IT!

That was EXTREMELY difficult!

After giving up in despair for a while, I tried again but tried a completely different strategy and got;

"{...}" indicates subscript.

S, F ∈ ℕ{0} (note I have now changed ℕ{1} to ℕ{0} to make it even more general)

x ∈ ]0, 1[ , x ∈ ℝ{>0}

∫[0, x] X^S (1 – X)^F dX = F! ∑[r=0, F] (-1)^r x^(S + 1 + r) / ( (S + 1 + r) r! (F – r)! )

plus have also checked that with a computer program with a numerical method and the above seems to be correct for many significant figures for the many input values I tied so I am sure that is totally correct.

The summation requires an iteration to evaluate it and it would be even better to have a solution without a need of iteration but, this is more than good enough! (for my research and my book I am writing)

I suppose that, in this case, no algebraic solution exists without requiring some sort of iteration for evaluation?

Although I worked this equation out*specifically*for a new type of probability distribution I have invented (that formula is just a small part of it else it wouldn't make sense for that because it doesn't normalize!), It has occurred to me that that new equation above might have wider application and many useful applications? - 08 Sep '16 17:28

The normal notation for a closed interval (one containing its endpoints) is [a, b], the usual notation for an open interval is (a, b), you don't need back to front brackets. In any case since the endpoints of the interval form a zero measure set and you are integrating whether or not you include them doesn't matter.*Originally posted by humy***ARR I have finally GOT IT!**

That was EXTREMELY difficult!

After giving up in despair for a while, I tried again but tried a completely different strategy and got;

"{...}" indicates subscript.

S, F ∈ ℕ{0} (note I have now changed ℕ{1} to ℕ{0} to make it even more general)

x ∈ ]0, 1[ , x ∈ ℝ{>0}

∫[0, x] X^S (1 – X)^F dX = F! ∑[r=0, F] (-1)^r x^( ...[text shortened]... ed to me that that new equation above might have wider application and many useful applications?

Yes, I'd missed the (-1)^r factors. The expression looks about right. - 08 Sep '16 18:17

I often have seen ]a, b[ used as well as (a, b) and, for no particular reason, just arbitrarily chose to use the ]a, b[ ones because I had to consistently choose one or the other. So I take it that the (a, b) is for some reason is the 'preferred' interval notation in mathematics?*Originally posted by DeepThought***The normal notation for a closed interval (one containing its endpoints) is [a, b], the usual notation for an open interval is (a, b), you don't need back to front brackets. ...**

If you confirm so, I will iterate though my whole book and change it to that 'preferred' one. - 09 Sep '16 11:15 / 1 edit

I took a quick look at Wikipedia, it gives both notations and says they are both defined by the International standard ISO 31-11, so I wouldn't bother changing anything. I think in Physics the round bracket notation is more common and so I queried the reversed square bracket notation because I'd never seen it before.*Originally posted by humy***I often have seen ]a, b[ used as well as (a, b) and, for no particular reason, just arbitrarily chose to use the ]a, b[ ones because I had to consistently choose one or the other. So I take it that the (a, b) is for some reason is the 'preferred' interval notation in mathematics?**

If you confirm so, I will iterate though my whole book and change it to that 'preferred' one.

So far as I can tell from Wikipedia the round bracket notation is older, but also is sometimes used to indicate an ordered pair such as coordinates, so Bourbaki introduced the reversed square bracket notation some time after 1935. The problem is that it is sometimes used to mean the complement of an open interval. - 09 Sep '16 12:02

I learnt the (a, b) one at university and this is the first time I have seen the other one. They both make intuitive sense and based on DeepThought's comments I would say pick the one you like best. Just be sure to explain/define your syntax the first time you use it in your book in case your readers are not familiar with the one you choose.*Originally posted by humy***I often have seen ]a, b[ used as well as (a, b) and, for no particular reason, just arbitrarily chose to use the ]a, b[ ones because I had to consistently choose one or the other. So I take it that the (a, b) is for some reason is the 'preferred' interval notation in mathematics?**

If you confirm so, I will iterate though my whole book and change it to that 'preferred' one. - 09 Sep '16 12:14

Can you graph that result? I wondered what it would look like graphed. Really want to see the finished book!*Originally posted by humy***ARR I have finally GOT IT!**

That was EXTREMELY difficult!

After giving up in despair for a while, I tried again but tried a completely different strategy and got;

"{...}" indicates subscript.

S, F ∈ ℕ{0} (note I have now changed ℕ{1} to ℕ{0} to make it even more general)

x ∈ ]0, 1[ , x ∈ ℝ{>0}

∫[0, x] X^S (1 – X)^F dX = F! ∑[r=0, F] (-1)^r x^( ...[text shortened]... ed to me that that new equation above might have wider application and many useful applications? - 09 Sep '16 17:09 / 1 edit

Given the large number of variables, it would either have to be a weird multidimensional plot, or he would need to fix some of the variables.*Originally posted by sonhouse***Can you graph that result?**

Having said that, the Beta function I believe is a special case of it which is graphed on Wikipedia:

https://en.wikipedia.org/wiki/Beta_function