An integral indexed by two partitions that mysteriously vanishes Let $\alpha,\beta\vdash n$ and define the polynomial
$$f_{\alpha,\beta}(x)=\sum_{\lambda \vdash n}\chi_\lambda(n)\chi_\lambda(\alpha)\chi_\lambda(\beta)x^{\ell(\lambda)-1},$$
where $\chi_\lambda$ are irreducible characters of the permutation group and $\ell$ is the length of a partition.
I am interested in the integral
$$I_{\alpha,\beta}=\int_0^\infty \frac{f_{\alpha,\beta}(x)}{(1+x)^{n+1}}dx,$$
and I have observed with surprise that $I_{\alpha,\beta}=0$ if $\ell(\alpha)+\ell(\beta)>n+1$.
How to prove this?
I know that $\chi_\lambda(n)=0$ unless $\lambda$ is a hook, i.e. of the form $(n-k,1^k)$, and that there is a formula for $\chi_\lambda$ when $\lambda$ is a hook. But I was not able to use that effectively (notice that $f_{\alpha,\beta}$ is never zero, only the integral is zero).
Another nice observation is that $f_{(1^n),(1^n)}(1)=0$ if $n$ is even and equals $(-1)^k{2k\choose k}$ if $n=2k-1$. Why?
Oh, I see. The integral is not relevant because it can be done to give
$$I_{\alpha\beta}=\frac{1}{n!}\sum_{\lambda \vdash n}\chi_\lambda(n)\chi_\lambda(\alpha)\chi_\lambda(\beta)(n-\ell(\lambda)-1)!(\ell(\lambda)-1)!$$or
$$I_{\alpha\beta}=\frac{1}{n}\sum_{\lambda \vdash n}\frac{1}{\chi_\lambda(1^n)}\chi_\lambda(n)\chi_\lambda(\alpha)\chi_\lambda(\beta),$$ and this is proportional to the number of factorizations of the full cycle with one factor in the conjugacy class $\alpha$ and the other in class $\beta$.
 A: Your nice observation is fairly easily explained: as you say $\chi_\lambda(n) = 0$ unless $\lambda$ is a hook partition of the form $(n-r,1^r)$ for some $r$. In this case, by the Murnaghan–Nakayama rule, we have $\chi_{(n-r,1^r)}(n) = (-1)^r$. Since there are $\binom{n-1}{r}$ standard Young tableaux of shape $(n-r,1^r)$, we have $\chi_{(n-r,1^r)}(1^n) = \binom{n-1}{r}$. Therefore
$$f_{(1^n),(1^n)}(1) = \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r}^2. $$
The right-hand side counts pairs $(X,Y)$ of subsets of $\{1,\ldots, n-1\}$ such that $|X|+|Y| = n-1$, weighted by $(-1)^{|X|}$. Let $X \circ Y = (X \backslash Y) \cup (Y \backslash X)$ be the symmetric difference of $X$ and $Y$. Moving the maximum element in $X \circ Y$ to the opposite set (so from $X$ to $Y$ if it is in $X$, and from $Y$ to $X$ if it is in $Y$) defines a sign-reversing involution on those pairs $(X,Y)$ with $X \not= Y$. The contribution from such pairs to the sum is therefore zero. We can have $X = Y$ only if $n-1 = 2k$ is even and $|X|=|Y| = k$; in this case the number of such pairs is $\binom{n-1}{k} = \binom{2k}{k}$, with sign $(-1)^k$. (I think you omitted the sign in your question.)
