# Integral: $\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t$

How to prove the following:

$$\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t=\frac{\pi}{2}\left(\ln(2)+\frac{\pi}{2} i\right)$$

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This doesn't make sense for $t \leq 1$ since you are taking the log of a negative number, unless you are specifying a certain branch cut. – nayrb Aug 24 '12 at 2:44
Are you trying to evaluate a real integral of a complex non-real function? Perhaps it was meant to be $\,\log|t+1|\,$? – DonAntonio Aug 24 '12 at 2:57

Deform the contour to pick up the residue at $t=i$, $$\begin{eqnarray*} \int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t &=& 2\pi i\, \mathrm{Res}_{t=i} \frac{\ln(t+1)}{t^2+1} \\ &=& 2\pi i \frac{\ln(1+i)}{2i} \\ &=& \frac{\pi}{2}\left(\ln 2+\frac{\pi}{2} i\right). \end{eqnarray*}$$ Here we have assumed the cut starting at $t=-1$ is slightly below the original contour. This appears to be the intended assumption. (If we instead assume the cut is slightly above the original contour we pick up the residue at $t=-i$ and get the complex conjugate of the quoted result.)

Note that the contribution at infinity vanishes since it goes like $\ln(R)/R\,(R\to\infty)$.

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A delightful answer! (+1) :D – user 1618033 Aug 30 '12 at 6:35

Let's evaluate the real and imaginary parts of the integral separately. Using $\Re(\ln(t+1)) = \ln(|t+1|)$ and $\Im(\ln(t+1)) = \pi [t < -1]$, where $[t<1]$ is the Iverson bracket: $$\begin{eqnarray} \Re \int_{-\infty}^\infty \frac{\ln(t+1)}{t^2+1} \mathrm{d} t &=& \int_{-\infty}^\infty \frac{\ln(|t+1|)}{t^2+1} \mathrm{d} t = \frac{1}{2} \int_{-\infty}^\infty \frac{\ln((t+1)^2 )}{t^2+1} \mathrm{d} t\\ \Im \int_{-\infty}^\infty \frac{\ln(t+1)}{t^2+1} \mathrm{d} t &=& \int_{-\infty}^{-1} \frac{\pi}{t^2+1} \mathrm{d} t \end{eqnarray}$$ The integral for the imaginary part is easy, since $\int \frac{\mathrm{d} t}{t^2+1} = \arctan(t) + C$: $$\int_{-\infty}^{-1} \frac{1}{t^2+1} \mathrm{d} t = \left.\arctan(t)\right|_{-\infty}^{-1} = \arctan(-1) - \lim_{t\to -\infty} \arctan(t) = -\frac{\pi}{4} - \left(-\frac{\pi}{2}\right) = \frac{\pi}{4}$$ Now to the real part: $$\int_{-\infty}^\infty \frac{\frac{1}{2} \log((t+1)^2)}{t^2+1} \mathrm{d} t = \int_{-\infty}^\infty \frac{\frac{1}{2} \log(t^2)}{t^2-2t +2} \mathrm{d} t = \int_0^\infty \log(t^2) \frac{1}{2} \left( \frac{1}{(t-1)^2+1}+\frac{1}{(t+1)^2 + 1} \right) \mathrm{d} t = \int_0^\infty \log(t^2) \frac{t^2+2}{t^4+4} \mathrm{d} t \stackrel{u=t^2}{=} \int_0^\infty \frac{\log(u)}{2} \frac{u+2}{u^2+4} \frac{\mathrm{d} u}{\sqrt{u}} = \frac{1}{2} \left.\frac{\mathrm{d} }{\mathrm{d} s} \int_0^\infty u^{s-1} \frac{u+2}{u^2 + 4} \mathrm{d} u \right|_{s=\frac{1}{2}}$$ The latter integral is a sum of Mellin-Barnes transforms of $\frac{1}{u^2+4}$: $$\int_0^\infty u^{s-1} \frac{u+2}{u^2 + 4} \mathrm{d} u = 2 \int_0^\infty u^{s-1} \frac{1}{u^2 + 4} \mathrm{d} u + \int_0^\infty u^{s} \frac{1}{u^2 + 4} \mathrm{d} u = \frac{2^{s-2} \pi}{\sin\left(\frac{\pi s}{2}\right)} + \frac{2^{s-2} \pi}{\cos\left(\frac{\pi s}{2}\right)}$$ Differentiating, and setting $s=\frac{1}{2}$ yields the result: $$\int_{-\infty}^\infty \frac{\frac{1}{2} \log((t+1)^2)}{t^2+1} \mathrm{d} t = \pi \log(2)$$ Recombining the real and imaginary parts: $$\int_{-\infty}^\infty \frac{\ln(t+1)}{t^2+1} \mathrm{d} t = \pi \log(2) + i \frac{\pi^2}{4}$$

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It would be easier if you define $$I(\alpha)=\int_{-\infty}^\infty\frac{\ln(\alpha t+1)}{t^2+1}dt.$$ Then \begin{eqnarray*} I'(\alpha)&=&\int_{-\infty}^\infty\frac{t}{(\alpha t+1)(t^2+1)}dt\\ &=&\frac{1}{\alpha^2+1}\int_{-\infty}^\infty\left(-\frac{1}{t+\frac{1}{\alpha}}+\frac{t}{t^2+1}+\frac{\alpha}{t^2+1}\right)dt\\ &=&\frac{1}{\alpha^2+1}\left(-\ln\frac{t+\frac{1}{\alpha}}{\sqrt{t^2+1}}+\alpha\arctan t\right)\big|_{-\infty}^\infty\\ &=&\frac{1}{\alpha^2+1}\left[\ln(-1)+\alpha\pi\right]\\ &=&\frac{\pi}{\alpha^2+1}(i+\alpha), \end{eqnarray*} and hence $I(\alpha)=\pi i\arctan\alpha+\frac{\pi}{2}\ln(\alpha^2+1)+C$. But $I(0)=0$ implies $C=0$. Thus $$I(1)=\pi i\frac{\pi}{4}+\frac{\pi}{2}\ln 2=\frac{\pi}{2}\left(\ln 2+i\frac{\pi}{2}\right).$$

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A related problem. Making the change of variables $t=u-1$ yields, $$\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t = \int _{-\infty }^{0 }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} + \int _{0 }^{\infty }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du}$$

The last integral has no problem, since $\ln(u)$ is real. For the other integral, we use the substitution $u=-z$, $$\int _{-\infty }^{0 }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} = \int _{0}^{\infty }\!{\frac {\ln \left( z \right)+\ln(-1) }{2+{z}^{2}+2\,z}}dz$$ That implies that

$$\int _{-\infty }^{\infty }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} = \int _{0 }^{\infty }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} + \int _{0}^{\infty }\!{\frac {\ln \left( z \right) }{2+{z}^{2}+2\,z}}dz +\int _{0}^{\infty }\!{\frac {\ln(-1) }{2+{z}^{2}+2\,z}}dz$$ $$= \frac{3\pi\ln(2)}{8} + \frac{\pi\ln(2)}{8}+\ln(-1)\,\frac{\pi}{4}$$ Now using $\ln(-1)=i\pi$ the result follow $$\frac{\pi \ln(2)}{2} + \frac{i\pi^2}{4} = \frac{\pi}{2}\left(\ln(2)+\frac{\pi}{2} i\right)$$

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How did you move from this $\int _{-\infty }^{\infty }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} = \int _{0 }^{\infty }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} + \int _{0}^{\infty }\!{\frac {\ln \left( z \right) }{2+{z}^{2}+2\,z}}dz +\int _{0}^{\infty }\!{\frac {\ln(-1) }{2+{z}^{2}+2\,z}}dz$ to this $= \frac{3\pi\ln(2)}{8} + \frac{\pi\ln(2)}{8}+\ln(-1)\,\frac{\pi}{4}$? – Frank Aug 24 '12 at 18:50
@MohammedAl-mubark: In the step where I applied the transformation $u=-z$, you get $\ln(u)=\ln(-z)=\ln(-1)+\ln(z)$ by the properties of the $\ln$ function. – Mhenni Benghorbal Aug 25 '12 at 3:26
But how did you get the result of $\int _{0}^{\infty }\!{\frac {\ln \left( z \right) }{2+{z}^{2}+2\,z}}dz +\int _{0}^{\infty }\!{\frac {\ln(-1) }{2+{z}^{2}+2\,z}}dz$? – Frank Aug 25 '12 at 11:31
@MohammedAl-mubark:See here; – Mhenni Benghorbal Aug 30 '12 at 17:09