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The solution was obtained already by Marchkenko and Pastur, in terms of the Cauchy transform $g(z)$ of the spectral density of $D+XX^{\rm T}$, see for example equations 2 + 3 of Spectrum of deformed random matrices and free probability:

$$g(z)=\int\frac{1}{z-t-(m/n)(z-t)g(z)-1+m/n}\rho(t)dt$$

where $\rho(t)$ is the spectral density of $D$. This holds for any random Hermitian perturbation $D$, irrespective of whether it is diagonal or not.

Let me try an alternative approach; The Cauchy transform of the Marchenko-Pastur distribution, supported on $[(1-\sqrt{\lambda})^2,(1+\sqrt{\lambda})^2]$, is given by $$g_{\rm MP}(z)=\frac{-\sqrt{(-\lambda+z-1)^2-4 \lambda}-\lambda+z+1}{2 z}.$$ The perturbation $D$, with $\rho(t)=w\delta(t-1)+(1-w)\delta(t)$, has Cauchy transform $$g_D(z)=\frac{w}{z-1}+\frac{1-w}{z}.$$ The Cauchy transform $g_{D+{\rm MP}}(z)$ of $D+XX^{\rm T}$ is given by the composition of free probability, $$R_{D+{\rm MP}}=R_{\rm MP}+R_D,\;\;R(z)=z-1/g(z).$$ This gives the Cauchy transform $$g_{D+{\rm MP}}(z)=\frac{1}{z}\left[ -\frac{2}{\sqrt{(\lambda-z+1)^2-4 \lambda}+\lambda-z-1}+\frac{z-1}{w+z-1}-1\right]^{-1}.$$

Now the goal is to integrate $\langle\log |t|\rangle_t=\int \rho_{D+{\rm MP}}(t)\log t\,dt$, which might be obtainable directly from the Cauchy transform via $$\frac{d}{dz}\langle\log |z-t|\rangle_t = \int \frac{\rho_{D+{\rm MP}}(t)}{z-t}\,dt=g_{D+{\rm MP}}(z).$$ The integration constant can be obtained from the large-$z$ limit, which should be just $\log z$.

The solution was obtained already by Marchkenko and Pastur, in terms of the Cauchy transform $g(z)$ of the spectral density of $D+XX^{\rm T}$, see for example equations 2 + 3 of Spectrum of deformed random matrices and free probability:

$$g(z)=\int\frac{1}{z-t-(m/n)(z-t)g(z)-1+m/n}\rho(t)\,dt$$

where $\rho(t)$ is the spectral density of $D$. This holds for any random Hermitian perturbation $D$, irrespective of whether it is diagonal or not.

Let me try an alternative approach; The Cauchy transform of the Marchenko-Pastur distribution, supported on $[(1-\sqrt{\lambda})^2,(1+\sqrt{\lambda})^2]$, is given by $$g_{\rm MP}(z)=\frac{-\sqrt{(-\lambda+z-1)^2-4 \lambda}-\lambda+z+1}{2 z}.$$ The perturbation $D$, with $\rho(t)=w\delta(t-1)+(1-w)\delta(t)$, has Cauchy transform $$g_D(z)=\frac{w}{z-1}+\frac{1-w}{z}.$$ The Cauchy transform $g_{D+{\rm MP}}(z)$ of $D+XX^{\rm T}$ is given by the composition of free probability, $$R_{D+{\rm MP}}=R_{\rm MP}+R_D,\;\;R(z)=z-1/g(z).$$ This gives the Cauchy transform $$g_{D+{\rm MP}}(z)=\frac{1}{z}\left[ -\frac{2}{\sqrt{(\lambda-z+1)^2-4 \lambda}+\lambda-z-1}+\frac{z-1}{w+z-1}-1\right]^{-1}.$$

Now the goal is to integrate $\langle\log |t|\rangle_t=\int \rho_{D+{\rm MP}}(t)\log t\,dt$, which might be obtainable directly from the Cauchy transform via $$\frac{d}{dz}\langle\log |z-t|\rangle_t = \int \frac{\rho_{D+{\rm MP}}(t)}{z-t} \, dt = g_{D+{\rm MP}}(z).$$ The integration constant can be obtained from the large-$z$ limit, which should be just $\log z$.

The solution was obtained already by Marchkenko and Pastur, in terms of the Cauchy transform $g(z)$ of the spectral density of $D+XX^{\rm T}$, see for example equations 2 + 3 of Spectrum of deformed random matrices and free probability:

$$g(z)=\int\frac{1}{z-t-(m/n)(z-t)g(z)-1+m/n}\rho(t)dt$$

where $\rho(t)$ is the spectral density of $D$. This holds for any random Hermitian perturbation $D$, irrespective of whether it is diagonal or not.

Let me try an alternative approach; The Cauchy transform of the Marchenko-Pastur distribution, supported on $[(1-\sqrt{\lambda})^2,(1+\sqrt{\lambda})^2]$, is given by $$g_{\rm MP}(z)=\frac{-\sqrt{(-\lambda+z-1)^2-4 \lambda}-\lambda+z+1}{2 z}.$$ The perturbation $D$, with $\rho(t)=w\delta(t-1)+(1-w)\delta(t)$, has Cauchy transform $$g_D(z)=\frac{w}{z-1}+\frac{1-w}{z}.$$ The Cauchy transform $g_{D+{\rm MP}}(z)$ of $D+XX^{\rm T}$ is given by the composition of free probability, $$R_{D+{\rm MP}}=R_{\rm MP}+R_D,\;\;R(z)=z-1/g(z).$$ This gives the Cauchy transform $$g_{D+{\rm MP}}(z)=\frac{1}{z}\left[ -\frac{2}{\sqrt{(\lambda-z+1)^2-4 \lambda}+\lambda-z-1}+\frac{z-1}{w+z-1}-1\right]^{-1}.$$

Now the goal is to integrate $\langle\log |t|\rangle_t=\int \rho_{D+{\rm MP}}(t)\log t\,dt$, which might be obtainable directly from the Cauchy transform via $$\frac{d}{dz}\langle\log |z-t|\rangle_t = \int \frac{\rho_{D+{\rm MP}}(t)}{z-t}\,dt=g_{D+{\rm MP}}(z).$$

The solution was obtained already by Marchkenko and Pastur, in terms of the Cauchy transform $g(z)$ of the spectral density of $D+XX^{\rm T}$, see for example equations 2 + 3 of Spectrum of deformed random matrices and free probability:

$$g(z)=\int\frac{1}{z-t-(m/n)(z-t)g(z)-1+m/n}\rho(t)dt$$

where $\rho(t)$ is the spectral density of $D$. This holds for any random Hermitian perturbation $D$, irrespective of whether it is diagonal or not.

Let me try an alternative approach; The Cauchy transform of the Marchenko-Pastur distribution, supported on $[(1-\sqrt{\lambda})^2,(1+\sqrt{\lambda})^2]$, is given by $$g_{\rm MP}(z)=\frac{-\sqrt{(-\lambda+z-1)^2-4 \lambda}-\lambda+z+1}{2 z}.$$ The perturbation $D$, with $\rho(t)=w\delta(t-1)+(1-w)\delta(t)$, has Cauchy transform $$g_D(z)=\frac{w}{z-1}+\frac{1-w}{z}.$$ The Cauchy transform $g_{D+{\rm MP}}(z)$ of $D+XX^{\rm T}$ is given by the composition of free probability, $$R_{D+{\rm MP}}=R_{\rm MP}+R_D,\;\;R(z)=z-1/g(z).$$ This gives the Cauchy transform $$g_{D+{\rm MP}}(z)=\frac{1}{z}\left[ -\frac{2}{\sqrt{(\lambda-z+1)^2-4 \lambda}+\lambda-z-1}+\frac{z-1}{w+z-1}-1\right]^{-1}.$$

Now the goal is to integrate $\langle\log |t|\rangle_t=\int \rho_{D+{\rm MP}}(t)\log t\,dt$, which might be obtainable directly from the Cauchy transform via $$\frac{d}{dz}\langle\log |z-t|\rangle_t = \int \frac{\rho_{D+{\rm MP}}(t)}{z-t}\,dt=g_{D+{\rm MP}}(z).$$ The integration constant can be obtained from the large-$z$ limit, which should be just $\log z$.

The solution was obtained already by Marchkenko and Pastur, in terms of the Cauchy transform $g(z)$ of the spectral density of $D+XX^{\rm T}$, see for example equations 2 + 3 of Spectrum of deformed random matrices and free probability:

$$g(z)=\int\frac{1}{z-t-(m/n)(z-t)g(z)-1+m/n}\rho(t)dt$$

where $\rho(t)$ is the spectral density of $D$. This holds for any random Hermitian perturbation $D$, irrespective of whether it is diagonal or not.

Let me try an alternative approach; The Cauchy transform of the Marchenko-Pastur distribution, supported on $[(1-\sqrt{\lambda})^2,(1+\sqrt{\lambda})^2]$, is given by $$g_{\rm MP}(z)=\frac{-\sqrt{(-\lambda+z-1)^2-4 \lambda}-\lambda+z+1}{2 z}.$$ The perturbation $D$, with $\rho(t)=w\delta(t-1)+(1-w)\delta(t)$, has Cauchy transform $$g_D(z)=\frac{w}{z-1}+\frac{1-w}{z}.$$ The Cauchy transform $g_{D+{\rm MP}}(z)$ of $D+XX^{\rm T}$ is given by the composition of free probability, $$R_{D+{\rm MP}}=R_{\rm MP}+R_D,\;\;R(z)=1/G(z)-1/z.$$ This gives the Cauchy transform $$g_{D+{\rm MP}}(z)=\left[\frac{1}{2} \left(\sqrt{(\lambda-z+1)^2-4 \lambda}-\lambda+z+1\right)+\frac{(z-1) z}{w+z-1}-\frac{1}{z}\right]^{-1}.$$

Now the goal is to integrate $\langle\log |t|\rangle_t=\int \rho_{D+{\rm MP}}(t)\log t\,dt$, which might be obtainable directly from the Cauchy transform via $$\frac{d}{dz}\langle\log |z-t|\rangle_t = \int \frac{\rho_{D+{\rm MP}}(t)}{z-t}\,dt=g_{D+{\rm MP}}(z).$$

The solution was obtained already by Marchkenko and Pastur, in terms of the Cauchy transform $g(z)$ of the spectral density of $D+XX^{\rm T}$, see for example equations 2 + 3 of Spectrum of deformed random matrices and free probability:

$$g(z)=\int\frac{1}{z-t-(m/n)(z-t)g(z)-1+m/n}\rho(t)dt$$

where $\rho(t)$ is the spectral density of $D$. This holds for any random Hermitian perturbation $D$, irrespective of whether it is diagonal or not.

Let me try an alternative approach; The Cauchy transform of the Marchenko-Pastur distribution, supported on $[(1-\sqrt{\lambda})^2,(1+\sqrt{\lambda})^2]$, is given by $$g_{\rm MP}(z)=\frac{-\sqrt{(-\lambda+z-1)^2-4 \lambda}-\lambda+z+1}{2 z}.$$ The perturbation $D$, with $\rho(t)=w\delta(t-1)+(1-w)\delta(t)$, has Cauchy transform $$g_D(z)=\frac{w}{z-1}+\frac{1-w}{z}.$$ The Cauchy transform $g_{D+{\rm MP}}(z)$ of $D+XX^{\rm T}$ is given by the composition of free probability, $$R_{D+{\rm MP}}=R_{\rm MP}+R_D,\;\;R(z)=z-1/g(z).$$ This gives the Cauchy transform $$g_{D+{\rm MP}}(z)=\frac{1}{z}\left[ -\frac{2}{\sqrt{(\lambda-z+1)^2-4 \lambda}+\lambda-z-1}+\frac{z-1}{w+z-1}-1\right]^{-1}.$$

Now the goal is to integrate $\langle\log |t|\rangle_t=\int \rho_{D+{\rm MP}}(t)\log t\,dt$, which might be obtainable directly from the Cauchy transform via $$\frac{d}{dz}\langle\log |z-t|\rangle_t = \int \frac{\rho_{D+{\rm MP}}(t)}{z-t}\,dt=g_{D+{\rm MP}}(z).$$