Integral of product of two Bernoulli polynomials I found formula of integral of product of two Bernoulli polynomial in Takashi Agoh & Karl Dilcher (http://www.sciencedirect.com/science/article/pii/S0022247X1100312X)
$\int_{0}^{1}B_k(t)B_m(t)dt=(-1)^{k-1}\frac{k!m!}{(k+m)!}B_{k+m}$
valid for $k+m\geq2$
where $B_k(t)$ and $B_m(t)$ are Bernoulli polynomials and $B_{k+m}$ is Bernoulli number
why it was valid just for $k+m\geq2$ not for $k+m<2$? 
Thanks for any helps
 A: Hint: try it for $k = 0, m = 1$.
As for "why didn't the proof work in this situation?", it's probably an inductive proof, and the base cases are probably $(m, k) = (2,0), (1,1)$ and $(0, 2)$. Sometimes that happens.  
Also: your formula, as written, doesn't quite make sense -- on the left you have a number, a definite integral; on the right, you have a function $B_{k+m}$. (Resolved: According to the comments, the same symbol is used for the name of the function as for its value at $x = 0$. Sigh. God help the programmer who has to write a program to handle this...)
OK, I'll do the case $k = 0, m = 1$. In this case, $B_k = 1$ and $B_m(x) = x - \frac{1}{2}$. $B_{m+k} = B_{m+k}(0) = B_1(0) = -\frac{1}{2}$. 
Now let's compute:
\begin{align}
\int_0^1 B_0(t) B_1(t) ~ dt &= \int_0^1 1 \cdot (x - \frac{1}{2})~dt \\
 &= \int_0^1 x - \frac{1}{2}~dt \\
 &= \left. \frac{x^2}{2} - \frac{x}{2} \right|_0^1 \\
&= 0. 
\end{align}
So...the reason the theorem doesn't work in that case is that $-1/2 \ne 0$. 
Why does the proof work in the other cases? Well, I'm pretty sure you can look up the proof elsewhere; there's not much reason to duplicate it here. 
