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I'm doing my IB Math HL IA on Bernoulli Polynomials, but what's bothering me is that I can't seem to prove the third definition, which is

$$\int_{0}^{1}B_n(x)dx=0$$

Can anyone help me?

user577215664
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Ilseok Jung
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    I'm not sure $\int_0^1 B_n(x) dx = 0$ is really a definition of the Bernoulli polynomials, maybe you copied the wrong image. If you could use MathJax that would also be appreciated. – Ian Oct 19 '17 at 16:21
  • On my math textbook, there are three definitions, or rather properties of the Bernoulli polynomials, the third one being $$\int_0^1 B_n(x)dx= 0 \tag{for n>1}$$. The first two are: $$\B_n'(x) =B_{n−1}(x)$$ and $$\B_0(x) = 1$$ – Ilseok Jung Oct 19 '17 at 16:30
  • $B_0(x)=1$ and $B_n'(x)=B_{n-1}(x)$ for $n \geq 1$ do not define $B_n$ uniquely because they leave all these unknown integration constants. For example, this definition only tells you $B_1(x)=x+C$ for some $C$. Therefore, to prove $\int_0^1 B_n(x) dx = 0$ for $n \geq 1$, you will need to add another rule to your definition. It is possible that your book is taking that rule to be $\int_0^1 B_n(x) dx = 0$ for $n \geq 1$ by definition, in which case there is nothing to prove. But I can't tell without some of the content of your book in front of me. – Ian Oct 19 '17 at 16:32
  • Also, your differential equation rule is not consistent with the definition of the Bernoulli numbers given on Wikipedia, which actually has $B_n'(x)=nB_{n-1}(x)$. – Ian Oct 19 '17 at 16:36
  • Thanks, but what really confuses me is that the definition in question is a definite integral. From what I learned, I only know that in a definite integration, the constant is not necessary as it will be cancelled out in the process of subtracting values. I also looked it up onWikipedia and I was surprised to find that the equations were different. I think the one on Wikipedia is similar to the bernoulli number which according to the book, is a completely new equation defined by $$b_n=n!B_n(0)$$ – Ilseok Jung Oct 19 '17 at 16:42
  • OK, so you actually do have a different convention, that's fine. The point here is that the integral condition specifies the constant of integration that you get from $B'n=B{n-1}$. For example, $B'_1=B_0=1$ so $B_1=x+c$; now the assumption $\int_0^1 B_1(x) dx = 0$ gives $1/2+c=0$ so $c=-1/2$. Note here that although you are right that the constant of integration cancels out when you take a definite integral, it doesn't cancel out when you take a definite integral of the antiderivative, which is what you're doing here. – Ian Oct 19 '17 at 16:57
  • ohh okay, I get it now. Thank you so much! – Ilseok Jung Oct 19 '17 at 16:58
  • OK. I'll make that an answer, it'd be appreciated if you accepted it. – Ian Oct 19 '17 at 17:00

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The definition of the Bernoulli polynomials that you are using is:

$$B_0(x)=1 \\ B_n(x)=B'_{n-1}(x) \quad n \geq 1 \\ \int_0^1 B_n(x) dx = 0 \quad n \geq 1.$$

(This is not quite the usual definition, which is $n!$ times this one, but that discrepancy is fine as long as we're aware of it.)

The third condition is necessary to specify the constants of integration that appear as you recursively apply the second condition. For example:

$$B_1(x)=\int B_0(x) dx = \int 1 dx = x+c_1 \\ \int_0^1 B_1(x) dx = \int_0^1 x+c_1 dx = \frac{1}{2} + c_1 = 0 \Rightarrow c_1=-1/2.$$

You could choose a different condition to specify the integration constants, if you wanted. For instance, $c_n=B_n(0)=\frac{b_n}{n!}$ where $b_n$ are the Bernoulli numbers (in the usual sense). But you do need something to specify them, otherwise you aren't really saying what polynomials you're considering.

Ian
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