Contour Integration of Inverse Hyperbolic Cosine

Question

I have the following integral

$$\int_1^\infty\frac{\text{arccosh}(x)}{x^2+1}\,dx$$

I am convinced that it can be done through contour integration, since we are integrating along the principle branch of the inverse hyperbolic cosine. But I am unsure of how to handle the inverse hyperbolic cosine when I'm trying to collect the real valued integral along the branch cut. My contour would be the keyhole contour that has a circle indented at $z = 1$, and we have the two paths that exist along the branch cuts which are at the negative real axis.

You should have residues at $z = i$ and $z=-i$, and the integrals around the small and large circles should be zero, but if anyone has any further thoughts on this integral or any other ways to go about it, I would love to see it.

Answer 1

I figured it out, this is a really cool problem that requires you to be very careful with the integration paths along the branch cuts.

Firstly, take the branch cut along the principal branch, this would be $(1,\infty)$. Now we consider the contour integral over $C$ of the function $\newcommand{arcosh}{\operatorname{arcosh}} \dfrac{\arcosh^2(z)}{z^2+1}$.

The Paths of C are broken up in the following way

$\oint_C = \int_{C_{1}}+\int_{C_{2}}+\int_{C_{3}}+\int_{C_{4}}+\int_{C_{5}}+\int_{C_{6}}+\int_{C_{\Gamma}}+\int_{C_{\gamma}}$

The integral around $C_{\Gamma}$ is the usual large circle $z = Re^{i\theta}$.

The integral around $C_{\gamma}$ is the indented circle $z = 1+\epsilon e^{i\theta}$, where we indent around $z = 1$.

You can use the fact that $|zf(z)| \rightarrow 0$ as $z \rightarrow \infty$, and that $(z-1)f(z) \rightarrow 0$ as $z \rightarrow 1$, to estimate both of those integrals to be $0$.

So all we need to do is consider each path very carefully, as the $\arcosh(z)$ is multi-valued across the branch cut. I've taken $\epsilon \rightarrow 0$ as a consequence of the integral along the indented circle to be zero. The reason I broke up $C_{2}$ and $C_{3}$ is because this integral transitions from the negative real axis to the positive real axis, so our choice of parametrization has to reflect this. The same thing applies for $C_{4}$ and $C_{5}$.

• $C_{1}: z = -x$ , $x \in (R,1]$

• $C_{2}: z = -x$ , $x \in [1,0]$

• $C_{3}: z = x$ , $x \in [0,1]$

• $C_{4}: z = x$ , $x \in [1,0]$

• $C_{5}: z = -x$ , $x \in [0,1]$

• $C_{6}: z = -x$ , $x \in [1,R)$

Now Lets look at integrals $C_{2}$ through $C_{5}$. This specifically concerns the branch cut $z \in [-1,1]$ and we must be careful of how we define the $\arcosh(z)$ around each path. Generally speaking,

$$\arcosh(z) = \pm \ln\left(i\sqrt{1-z^{2}}+z\right), \qquad z \in [-1,1]$$

For $C_{2}$ and $C_{3}$ use the positive branch, for $C_{4}$ and $C_{5}$ use the negative branch.

$$\int_{C_{2}}\frac{\arcosh^2(z)}{z^2+1}dz+\int_{C_{3}}\frac{\arcosh^2(z)}{z^2+1}dz+\int_{C_{4}}\frac{\arcosh^2(z)}{z^2+1}dz+\int_{C_{5}}\frac{\arcosh^2(z)}{z^2+1}dz.$$

To be continued I need sleep!

Edit: It was a long nap!!

$\int_{C_{2}}\frac{\left(\ln\left(i\sqrt{1-z^{2}}+z\right)\right)^2}{z^2+1}dz+\int_{C_{3}}\frac{\left(\ln\left(i\sqrt{1-z^{2}}+z\right)\right)^2}{z^2+1}dz+\int_{C_{4}}\frac{\left(-\ln\left(i\sqrt{1-z^{2}}+z\right)\right)^2}{z^2+1}dz+\int_{C_{5}}\frac{\left(-\ln\left(i\sqrt{1-z^{2}}+z\right)\right)^2}{z^2+1}dz$ =

$\int_1^0\frac{\ln^2\left(i\sqrt{1-x^{2}}-x\right)}{x^2+1}(-dx)+\int_0^1\frac{\ln^2\left(i\sqrt{1-x^{2}}+x\right)}{x^2+1}dx+\int_1^0\frac{\ln^2\left(i\sqrt{1-x^{2}}+x\right)}{x^2+1}dx+\int_0^1\frac{\ln^2\left(i\sqrt{1-z^{2}}-x\right)}{x^2+1}(-dx)$ =

$\int_0^1\frac{\ln^2\left(i\sqrt{1-x^{2}}-x\right)}{x^2+1}dx+\int_0^1\frac{\ln^2\left(i\sqrt{1-x^{2}}+x\right)}{x^2+1}dx-\int_0^1\frac{\ln^2\left(i\sqrt{1-x^{2}}+x\right)}{x^2+1}dx-\int_0^1\frac{\ln^2\left(i\sqrt{1-x^{2}}-x\right)}{x^2+1}dx$

But all of these terms cancel and the contribution of these integrals is zero. So now we consider the integrals around $C_{1}$ and $C_{6}$. The $\arcosh(z)$ has the following values on these integrals.

$\arcosh(z) = \pm i\pi+\ln\left(\sqrt{z^{2}-1}-z\right)$, $z \in (-\infty,-1]$

For $C_{1}$ use the positive branch, for $C_{6}$ use the negative branch.

$\int_{C_{1}}\frac{\arcosh^2(z)}{z^2+1}dz+\int_{C_{6}}\frac{\arcosh^2(z)}{z^2+1}dz$

$=\int_{C_{1}}\frac{\left(i\pi+\ln\left(\sqrt{z^{2}-1}-z\right)\right)^2}{z^2+1}dz+\int_{C_{6}}\frac{\left(-i\pi+\ln\left(\sqrt{z^{2}-1}-z\right)\right)^2}{z^2+1}dz$

$=\int_R^1\frac{\left(i\pi+\ln\left(\sqrt{x^{2}-1}+x\right)\right)^2}{x^2+1}(-dx)+\int_1^R\frac{\left(-i\pi+\ln\left(\sqrt{x^{2}-1}+x\right)\right)^2}{x^2+1}(-dx)$ =

$=\int_1^R\frac{-\pi^{2}+2\pi i\ln\left(\sqrt{x^{2}-1}+x\right)+\ln^2\left(\sqrt{x^{2}-1}+x\right)}{x^2+1}dx+\int_1^R\frac{\pi^{2}+2\pi i\ln\left(\sqrt{x^{2}-1}+x\right)-\ln^2\left(\sqrt{x^{2}-1}+x\right)}{x^2+1}dx$

$=\int_1^R\frac{-\pi^{2}+2\pi i\ln\left(\sqrt{x^{2}-1}+x\right)+\ln^2\left(\sqrt{x^{2}-1}+x\right)+\pi^{2}+2\pi i\ln\left(\sqrt{x^{2}-1}+x\right)-\ln^2\left(\sqrt{x^{2}-1}+x\right)}{x^2+1}dx$

Canceling the appropriate terms, and letting $R \rightarrow \infty$

$$4 \pi i \int_1^\infty \frac{\ln\left(\sqrt{x^2-1}+x\right)}{x^2+1}dx = 4 \pi i \int_1^\infty \frac{\arcosh(x)}{x^2+1}dx = 4 \pi i I$$

where $I$ is the integral we are looking for. Now by construction this is equal to

$$\oint_C \frac{\arcosh^2(z)}{z^2+1}dz$$

and this is equal to the residues that are collected at $z = \pm i$, multiplied by a factor of $2 \pi i$. A quick computation shows,

$$\oint_C \frac{\arcosh^2(z)}{z^2+1}dz = 4 \pi i I = 2 \pi^2 i \ln\left(\sqrt{2}+1\right)$$

In conclusion,

$$I = \frac{\pi}{2} \ln\left(\sqrt{2}+1\right) = \frac{\pi}{2} \operatorname{arsinh}(1)$$

Answer 2

We can circumvent consideration of branch cuts by leveraging Fubini's theorem $(*)$:

$$\begin{align*} I = \int_1^\infty \frac{\operatorname{arcosh}x}{1+x^2} \, dx &= \int_1^\infty \int_1^x \frac{dt\,dx}{\left(1+x^2\right)\sqrt{t^2-1}} \\ &\stackrel{(*)}= \int_1^\infty \int_t^\infty \frac{dx\,dt}{\left(1+x^2\right)\sqrt{t^2-1}} \\ &\stackrel{(*)}= \int_1^\infty \int_1^\infty \frac{t\,dt\,dx}{\left(1+t^2x^2\right)\sqrt{t^2-1}} & x\to tx \\ &= \frac12 \int_1^\infty \int_{-\infty}^\infty \frac{du\,dx}{1+x^2+x^2u^2} & u=t^2-1;\text{ symmetry} \end{align*}$$

The integral w.r.t. $u$ can be easily computed through the residue theorem using a semicircular contour enclosing a simple pole at $u=i\dfrac{\sqrt{1+x^2}}x$, with residue $\dfrac1{2ix\sqrt{1+x^2}}$, hence

$$I = \frac\pi2 \int_1^\infty \frac{dx}{x\sqrt{1+x^2}} \stackrel{x\to1/x}= \frac\pi2 \int_0^1 \frac{dx}{\sqrt{1+x^2}} = \boxed{\frac\pi2\operatorname{arsinh}1}$$