Average of the maximum matrix element over the Haar measure

The answer to the question as stated (maximum of row elements) has been solved in Extreme statistics of complex random and quantum chaotic states, see also this MO posting:

$$\int dU \max_j |U_{1,j}|^2 =\frac{1}{d}\sum_{j=1}^d \frac{1}{j}.$$

For large $d$ this tends to $(1/d)\log d$. The complete probability distribution of the row-maximum is known.

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The maximum of all matrix elements is more difficult and only large-$d$ asymptotics has a closed-form expression, see Maxima of entries of Haar distributed matrices (alternative link).

If $W_d$ is the maximum matrix element in absolute value of a Haar-distributed $d\times d$ unitary matrix, then $W_d^2\rightarrow (2/d)\log d$ in probability for $d\rightarrow\infty$, so twice as large as for the row-maximum.

$\newcommand{\C}{\mathbb C}$$\newcommand{\R}{\mathbb R}$Any linear isometry of $\C^d$ is a unitary transformation. Therefore, the distribution of the random vector $V:=(X_1,Y_1,\dots,X_d,Y_d)$ is uniform on the unit sphere in $\R^{2d}$, where $X_j:=\Re U_{1,j}$ and $Y_j:=\Im U_{1,j}$. So, $V$ equals $W:=(W_1,\dots,W_{2d})$ in distribution, where $$W_j:=\frac{Z_j}{\sqrt{Z_1^2+\dots+Z_{2d}^2}} $$ and $Z_1,\dots,Z_{2d}$ are iid standard normal random variables (r.v.'s). So, the random vector $(|U_{1,1}|^2 ,\dots,|U_{1,d}|^2)$ equals $R:=(R_1,\dots,R_d)$ in distribution, where $$R_j:=\frac{T_j}{T_1+\dots+T_d} $$ and $T_k:=Z_{2k-1}^2+Z_{2k}^2$, so that $T_1,\dots,T_d$ are iid standard exponential r.v.'s. So, $\max_j|U_{1,j}|^2$ equals $\max_j R_j$ in distribution. The distribution and, in particular, the expectation of $\max_j R_j$ were found a long time ago; see e.g. Irwin (1955) and historical references therein going back to as far as 1897. In particular, according to formula (15) in Cochran's paper, $$E\max_j|U_{1,j}|^2=E\max_j R_j=\frac1d\,\Big(1+\frac12+\dots+\frac1d\Big). $$