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I cannot figure out an integral (which involves certain approximations) in the textbook Quantum Field Theory by Peskin and Schroeder.

On P.220 Eq.(7.28-29), it is mentioned that the integral (7.28)

$$\delta m =\frac{\alpha}{2\pi}m_0\int_0^1 dx(2-x)\log\left(\frac{x\Lambda^2}{(1-x)^2m_0^2+x\mu^2}\right)$$

takes the form (7.29) when $\Lambda \rightarrow +\infty$:

$$\delta m \rightarrow \frac{3\alpha}{4\pi}m_0\log\left(\frac{\Lambda^2}{m_0^2}\right)$$

The thing is that it seems to me the integral is divergent at $x\rightarrow 0$ for finite $\Lambda$, then how to obtain (7.29) at the desired limit?

Any comment is really appreciated.

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Because $\mu$ is the mass parameter of photon, we can set $\mu=0$. Then we can see \begin{eqnarray*} \delta m & = & \frac{\alpha}{2\pi}m_{0}\int_{0}^{1}dx(2-x)\mathrm{ln}\bigg(\frac{x\Lambda^{2}}{(1-x)^{2}m_{0}^{2}}\bigg)\\ & = & \frac{\alpha}{2\pi}m_{0}\int_{0}^{1}dx(2-x)\bigg(\mathrm{ln}x-2\mathrm{ln}(1-x)+\mathrm{ln}\frac{\Lambda^{2}}{m_{0}^{2}}\bigg)\\ & = & \frac{3\alpha}{4\pi}m_{0}\mathrm{ln}\frac{\Lambda^{2}}{m_{0}^{2}}+\frac{\alpha}{2\pi}m_{0}\bigg(2\int_{0}^{1}dx\mathrm{ln}x-\int_{0}^{1}dxx\mathrm{ln}x\\ & & -2\int_{0}^{1}dx\mathrm{ln}(1-x)-2\int_{0}^{1}dx(1-x)\mathrm{ln}(1-x)\bigg)\\ & = & \frac{3\alpha}{4\pi}m_{0}\mathrm{ln}\frac{\Lambda^{2}}{m_{0}^{2}}+\frac{\alpha}{2\pi}m_{0}\bigg(2\int_{0}^{1}dx\mathrm{ln}x-\int_{0}^{1}dxx\mathrm{ln}x-2\int_{0}^{1}dx\mathrm{ln}x-2\int_{0}^{1}dxx\mathrm{ln}x\bigg)\\ & = & \frac{3\alpha}{4\pi}m_{0}\mathrm{ln}\frac{\Lambda^{2}}{m_{0}^{2}}-\frac{3\alpha}{2\pi}m_{0}\int_{0}^{1}dxx\mathrm{ln}x\\ & = & \frac{3\alpha}{4\pi}m_{0}\mathrm{ln}\frac{\Lambda^{2}}{m_{0}^{2}}-\frac{3\alpha}{2\pi}m_{0}\times(-\frac{1}{4})\\ & \rightarrow & \frac{3\alpha}{4\pi}m_{0}\mathrm{ln}\frac{\Lambda^{2}}{m_{0}^{2}}\ \ \text{as}\ \ \Lambda\rightarrow\infty \end{eqnarray*}

where we have used \begin{eqnarray*} \int_{0}^{1}dxx\mathrm{ln}x & = & \bigg(\frac{x^{2}}{2}\mathrm{ln}x\bigg)_{0}^{1}-\frac{1}{2}\int_{0}^{1}dxx=-\frac{1}{4} \end{eqnarray*}

There is no divergence in the final result, although the integrand is divergent at $x=0$ and $x=1$.

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Note that the integrand is logarithmic divergent when $x\to 0$, and that is fine as $\int \log x dx= x(\log x-1)\to 0 $ as $x\to 0$. Specifically, we have the leading behaviour $$\delta m \sim\frac{\alpha}{2\pi}m_0\int_0^1dx(2-x)\log(\frac{\Lambda^2}{m_0^2})=\frac{3\alpha}{4\pi}m_0\log\left(\frac{\Lambda^2}{m_0^2}\right).$$

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