A monster antiderivative $\int\ln(A\cosh(\alpha x)+B\sinh(\alpha x)+C)dx$

Question

After spending almost a week, I gave up on finding the exact form of this antiderivative:

$$\int \ln(A\cosh(\alpha x)+B\sinh(\alpha x)+C)dx$$

What are some nice (i.e. well written and compact) approximated form? While, using Taylor expansion is welcome, avoiding Euler formula is appreciated.

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BACKGROUND: I need to formulate the integral of the natural logarithm of a quantity which looks like $f(x)=A\cosh(\alpha x)+B\sinh(\alpha x)+C$. I need to come up with something compact and sufficiently accurate that I can later use it for any values of $A$, $B$, $C$ and $\alpha$. In other words, the exact solution seems to be too complicated to be used as a compact formula. (I know that the approximated form may depends on the ranges of values).

PS Even applying Taylor expansion (generated by wolfram) does not look fun: $$\ln(A + C) + \frac{B}{(A + C)}(ax) + \frac{(A^2 + A C - B^2)}{2 (A + C)^2}(ax)^2 + \frac{B(-2 A^2 - A C + 2 B^2 + C^2)}{6 (A + C)^3}(ax)^3 + \frac{(-2 A^4 - 3 A^3 C + 8 A^2 B^2 + A (4 B^2 C + C^3) - 6 B^4 - 4 B^2 C^2)}{24 (A + C)^4}(ax)^4 + \frac{B(16 A^4 + 19 A^3 C - A^2 (40 B^2 + 9 C^2) - A (20 B^2 C + 11 C^3) + 24 B^4 + 20 B^2 C^2 + C^4)}{120 (A + C)^5}(ax)^5 + O(x^6)$$

Answer 1

I will assume that you are only interested in real values. That is,

Assumption. $A$, $B$, $C$, $\alpha $ are all in $\mathbb{R}$.

In particular,

$$f(x) = A\cosh x + B\sinh x + C$$

is a real-valued function on $\mathbb{R}$. Then this setting allows us to avoid some pesky issue regarding the branch cut of the complex logarithm. So let us compute the integral.

• Let $p, q$ be zeros of the quadratic polynomial $\frac{A+B}{2}t^2 + Ct + \frac{A-B}{2}$. In other words, they satisfy the identity $ \frac{A+B}{2}t^2 + Ct + \frac{A-B}{2} = \frac{A+B}{2}(t - p)(t - q)$. Then plugging $t = e^{\alpha x}$, we find that

  $$f(x) = \frac{A+B}{2} e^{\alpha x} (1-pe^{-\alpha x}) (1-qe^{-\alpha x}) $$

  In particular, we have

  \begin{align*} \log |f(x)| = \log\left|\frac{A+B}{2}\right| + \alpha x + \log |1-pe^{-\alpha x}| + \log |1-qe^{-\alpha x}|. \tag{1} \end{align*}

  (Here, I am assuming that you are only interested in the region where $f(x) > 0$. Then taking absolute values causes no harm.)

• In order to integrate the above equation, we need a special function called the dilogarithm. It is the function which is denoted by $\operatorname{Li}_2$ and is defined by

  $$ \operatorname{Li}_2(z) = -\int_{0}^{z} \frac{\log(1-t)}{t} \, dt = -\int_{0}^{1} \frac{\log(1-zu)}{u} \, du, \quad z \in \mathbb{C}\setminus[1,\infty). $$

  (Here, the first integral is taken along the line segment joining $0$ and $z$. The equivalence of two integrals can be shown by the substitution $t = zu$.) So $\operatorname{Li}_2(z)$ is not differentiable along the branch cut $[1,\infty)$. On the other hand, its real part behaves much better since

  $$ \operatorname{Re}\operatorname{Li}_2(z) = -\int_{0}^{1} \frac{\log|1-zu|}{u} \, du \tag{2}$$

  A bit of complex analysis tells that both sides of $\text{(2)}$, understood as function of real variables $x$ and $y$ (with $z = x+iy$), extends to a smooth on all of $\mathbb{R}^2\setminus\{(1,0)\}$. As a useful consequences, for any $0 < a < b$ we have

  \begin{align*} -\int_{a}^{b} \frac{\log|1-zu|}{u} \, du &= -\int_{0}^{b} \frac{\log|1-zu|}{u} \, du - \left( -\int_{0}^{a} \frac{\log|1-zu|}{u} \, du \right) \\ &= -\int_{0}^{1} \frac{\log|1-bzv|}{v} \, dv - \left( -\int_{0}^{1} \frac{\log|1-azw|}{w} \, dw \right) \\ &= \operatorname{Re}\operatorname{Li}_2(bz) - \operatorname{Re}\operatorname{Li}_2(az). \end{align*}

  (Here, we utilized the substitution $u = bv$ and $u = aw$ for respective terms.) This is useful when computing some indefinite integrals to come.

• Using the previous computation, we can integrate $\text{(1)}$. Indeed, let us substitute $u=e^{-\alpha x}$. Then

  $$\int \log|1 - p e^{-\alpha x}| \, dx = -\frac{1}{\alpha}\int \frac{\log|1 - pu|}{u} \, du = \frac{1}{\alpha}\operatorname{Re}\operatorname{Li}_2(pu) + \text{constant} $$

  This remains true if we replace $p$ by $q$. Therefore we get

\begin{align*} &\int \log|f(x)| \, dx \\ &\hspace{1em}= x\log\left|\frac{A+B}{2}\right| + \frac{\alpha x^2}{2} + \frac{1}{\alpha}\operatorname{Re}\operatorname{Li}_2(pe^{-\alpha x}) + \frac{1}{\alpha}\operatorname{Re}\operatorname{Li}_2(qe^{-\alpha x}) + \text{constant}. \end{align*}

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Example. Let $A = -1$, $B = 2$, $C = 1$ and $\alpha = 3$. Then both computation shows that

$$ \int_{-2}^{3} \log |f(x)| \, dx \approx 17.88480462181253086\cdots. $$

The following is an actual computation using Mathematica 11: