Integral $\int_{-D}^D \frac{1}{x-ia} \log \frac{(x-2D)^2+a^2}{x^2+a^2} dx$

Question

For $a>0$ and $D>0$, I want to evaluate the integral $$\int_{-D}^D \frac{1}{x-ia} \log \frac{(x-2D)^2+a^2}{x^2+a^2} dx.$$ Here $i$ is the imaginary unit. This integral arises when evaluating a Feynman diagram in physics.

Answer 1

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HINT:

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Define the function $\mathcal{I}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{C}$ via the definite integral

$$\mathcal{I}{\left(a,D\right)}:=\int_{-D}^{D}\mathrm{d}x\,\frac{1}{x-ia}\ln{\left(\frac{\left(x-2D\right)^{2}+a^{2}}{x^{2}+a^{2}}\right)}.$$

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For any $\left(a,D\right)\in\mathbb{R}_{>0}^{2}$, we have

$$\begin{align} \mathcal{I}{\left(a,D\right)} &=\int_{-D}^{D}\mathrm{d}x\,\frac{1}{x-ia}\ln{\left(\frac{\left(x-2D\right)^{2}+a^{2}}{x^{2}+a^{2}}\right)}\\ &=\int_{-\frac{D}{a}}^{\frac{D}{a}}\mathrm{d}x\,\frac{1}{x-i}\ln{\left(\frac{\left(x-2\frac{D}{a}\right)^{2}+1}{x^{2}+1}\right)};~~~\small{\left[x\mapsto ax\right]}\\ &=\mathcal{I}{\left(1,\frac{D}{a}\right)}.\\ \end{align}$$

Thus, WLOG we may go ahead and assume $a=1$ in our general evaluation of $\mathcal{I}{\left(a,D\right)}$.

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Suppose $D\in\mathbb{R}_{>0}$. We obtain the following equivalent form:

$$\begin{align} \mathcal{I}{\left(1,D\right)} &=\int_{-D}^{D}\mathrm{d}x\,\frac{1}{x-i}\ln{\left(\frac{\left(x-2D\right)^{2}+1}{x^{2}+1}\right)}\\ &=\int_{-D}^{D}\mathrm{d}x\,\frac{x+i}{x^{2}+1}\ln{\left(\frac{\left(x-2D\right)^{2}+1}{x^{2}+1}\right)}\\ &=\int_{-2D}^{0}\mathrm{d}y\,\frac{\left(y+D\right)+i}{\left(y+D\right)^{2}+1}\ln{\left(\frac{\left(y-D\right)^{2}+1}{\left(y+D\right)^{2}+1}\right)};~~~\small{\left[x=y+D\right]}\\ &=\int_{0}^{2D}\mathrm{d}t\,\frac{\left(D-t\right)+i}{\left(t-D\right)^{2}+1}\ln{\left(\frac{\left(t+D\right)^{2}+1}{\left(t-D\right)^{2}+1}\right)};~~~\small{\left[y=-t\right]}\\ &=\int_{0}^{2D}\mathrm{d}t\,\frac{\left(t-D\right)-i}{\left(t-D\right)^{2}+1}\ln{\left(\frac{\left(t-D\right)^{2}+1}{\left(t+D\right)^{2}+1}\right)}.\\ \end{align}$$

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Given fixed but arbitrary $\left(p,q\right)\in\mathbb{R}^{2}$ such that $p\neq q$, consider the linear fractional transformation given implicitly by the relation

$$\left(t-q\right)\left(1+u\right)=\left(p-q\right).$$

Solving for either $t$ or $u$, we obtain

$$t=\frac{p+qu}{1+u}\land\frac{p-t}{t-q}=u.$$

Also observe that

$dt=du\,\frac{\left(-1\right)\left(p-q\right)}{\left(1+u\right)^{2}},$

$\left(t-D\right)=\frac{\left(p-D\right)+\left(q-D\right)u}{\left(1+u\right)},$

$\left(t-D\right)^{2}+1=\frac{\left[\left(p-D\right)^{2}+1\right]+2\left[\left(p-D\right)\left(q-D\right)+1\right]u+\left[\left(q-D\right)^{2}+1\right]u^{2}}{\left(1+u\right)^{2}},$

$\left(t+D\right)^{2}+1=\frac{\left[\left(p+D\right)^{2}+1\right]+2\left[\left(p+D\right)\left(q+D\right)+1\right]u+\left[\left(q+D\right)^{2}+1\right]u^{2}}{\left(1+u\right)^{2}}.$

Ideally, we'd like to choose values for $p$ and $q$ satisfying the following pair of equations:

$$\begin{cases} &\left(p-D\right)\left(q-D\right)+1=0,\\ &\left(p+D\right)\left(q+D\right)+1=0.\\ \end{cases}$$

Equivalently,

$$\begin{cases} &pq-D\left(p+q\right)+D^{2}+1=0\\ &pq+D\left(p+q\right)+D^{2}+1=0,\\ \end{cases}$$

$$\implies\begin{cases} &\left(p+q\right)=0\\ &pq+D^{2}+1=0,\\ \end{cases}$$

$$\implies\begin{cases} &p=\pm\sqrt{1+D^{2}},\\ &q=-p.\\ \end{cases}$$

We will use the unique solution $\left(p,q\right)\in\mathbb{R}^{2}$ such that $p>q$:

$$\left(p,q\right)=\left(\sqrt{1+D^{2}},-\sqrt{1+D^{2}}\right).$$

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Continuing with our evaluation of $\mathcal{I}$, we find

$$\begin{align} \mathcal{I}{\left(1,D\right)} &=\int_{0}^{2D}\mathrm{d}t\,\frac{\left(t-D\right)-i}{\left(t-D\right)^{2}+1}\ln{\left(\frac{\left(t-D\right)^{2}+1}{\left(t+D\right)^{2}+1}\right)}\\ &=\int_{-\frac{p}{q}}^{\frac{p-2D}{2D-q}}\mathrm{d}u\,\frac{\left(-1\right)\left(p-q\right)}{\left(1+u\right)^{2}}\left[\frac{\left(p-D\right)+\left(q-D\right)u}{\left(1+u\right)}-i\right]\\ &~~~~~\times\frac{\left(1+u\right)^{2}}{\left[\left(p-D\right)^{2}+1\right]+\left[\left(q-D\right)^{2}+1\right]u^{2}}\\ &~~~~~\times\ln{\left(\frac{\left[\left(p-D\right)^{2}+1\right]+\left[\left(q-D\right)^{2}+1\right]u^{2}}{\left[\left(p+D\right)^{2}+1\right]+\left[\left(q+D\right)^{2}+1\right]u^{2}}\right)};~~~\small{\left[t=\frac{p+qu}{1+u}\right]}\\ &=\int_{\frac{p-2D}{p+2D}}^{1}\mathrm{d}u\,\left[\frac{\left(p-D\right)-\left(p+D\right)u}{\left(1+u\right)}-i\right]\\ &~~~~~\times\frac{2p}{\left[\left(p-D\right)^{2}+1\right]+\left[\left(p+D\right)^{2}+1\right]u^{2}}\\ &~~~~~\times\ln{\left(\frac{\left[\left(p-D\right)^{2}+1\right]+\left[\left(p+D\right)^{2}+1\right]u^{2}}{\left[\left(p+D\right)^{2}+1\right]+\left[\left(p-D\right)^{2}+1\right]u^{2}}\right)}\\ &=\int_{\frac{p-2D}{p+2D}}^{1}\mathrm{d}u\,\left[\frac{\left(p-D\right)-\left(p+D\right)u}{\left(1+u\right)}-i\right]\\ &~~~~~\times\frac{2p}{2p\left(p-D\right)+2p\left(p+D\right)u^{2}}\\ &~~~~~\times\ln{\left(\frac{2p\left(p-D\right)+2p\left(p+D\right)u^{2}}{2p\left(p+D\right)+2p\left(p-D\right)u^{2}}\right)}\\ &=\int_{\frac{p-2D}{p+2D}}^{1}\mathrm{d}u\,\left[\frac{\left(p-D\right)-\left(p+D\right)u}{\left(1+u\right)}-i\right]\\ &~~~~~\times\frac{1}{\left(p-D\right)+\left(p+D\right)u^{2}}\ln{\left(\frac{\left(p-D\right)+\left(p+D\right)u^{2}}{\left(p+D\right)+\left(p-D\right)u^{2}}\right)}.\\ \end{align}$$

Setting $d:=\sqrt{1+D^{2}}-D\in\left(0,1\right)$ and $z:=\frac{\sqrt{1+D^{2}}-2D}{\sqrt{1+D^{2}}+2D}\in\left(-1,1\right)$, we then find

$$\begin{align} \mathcal{I}{\left(1,D\right)} &=\int_{\frac{p-2D}{p+2D}}^{1}\mathrm{d}u\,\left[\frac{\left(p-D\right)-\left(p+D\right)u}{\left(1+u\right)}-i\right]\\ &~~~~~\times\frac{1}{\left(p-D\right)+\left(p+D\right)u^{2}}\ln{\left(\frac{\left(p-D\right)+\left(p+D\right)u^{2}}{\left(p+D\right)+\left(p-D\right)u^{2}}\right)}\\ &=\int_{\frac{\sqrt{1+D^{2}}-2D}{\sqrt{1+D^{2}}+2D}}^{1}\mathrm{d}u\,\left[\frac{d-d^{-1}u}{\left(1+u\right)}-i\right]\frac{1}{d+d^{-1}u^{2}}\ln{\left(\frac{d+d^{-1}u^{2}}{d^{-1}+du^{2}}\right)}\\ &=\int_{z}^{1}\mathrm{d}u\,\left[\frac{d^{2}-u}{\left(1+u\right)}-id\right]\frac{1}{d^{2}+u^{2}}\ln{\left(\frac{d^{2}+u^{2}}{1+d^{2}u^{2}}\right)}.\\ \end{align}$$

This is as far as I was willing to work on the problem for the time being, but hopefully it's enough of a simplification for some to consider useful.

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Answer 2

It does not seem to be the most pleasant integral. Focusing on the antoderivative, let first $x=a y$ and $k=\frac{2D}a$to make $$I=\int \frac{1}{x-ia} \log \left(\frac{(x-2D)^2+a^2}{x^2+a^2}\right)\,dx=\int\frac{1}{y-i}\log \left(\frac{\left(y-k\right)^2+1}{y^2+1}\right)\,dy$$ Deloping the logarithm of the ratio, you then face two integrals looking like $$J_k=\int \frac{1}{y-i}\log \left({\left(y-k\right)^2+1}{}\right)\,dy$$ $$J_k=\text{Li}_2\left(\frac{k-y-i}{k-2 i}\right)+\text{Li}_2\left(\frac{k-y+i}{k}\right)+\log (y-i) \log \left((k-y)^2+1\right)-\log (k) \log (-k+y-i)+\left(\log \left(\frac{y-i}{k-2 i}\right)-\log (y-i)\right) \log (-k+y+i)$$ $$J_0=\text{Li}_2\left(-\frac{1}{2} i (y+i)\right)+\left(\log \left(y^2+1\right)-\log (y+i)\right) \log (y-i)-\frac{1}{2} \log ^2(y-i)+$$ $$\log \left(\frac{1}{2} (1+i y)\right) \log (y+i)$$

The definite integral will not be very nice looking.