Finding $ \int_0^x \frac{\sinh^{-1}\left(\alpha\tanh(kx)\right)}{\sinh^{-1}⁡(k)} dx $

Question

I'm trying to solve this equation but not quite sure which calculation method to use.

$$ \int_0^x \frac{\sinh^{-1}\left(\alpha\tanh(kx)\right)}{\sinh^{-1}⁡(k)} dx $$

where $\alpha$ and $k$ are just constants.

Answer 1

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

$$\mathcal{I}{\left(a,z\right)}:=\int_{0}^{z}\mathrm{d}\tau\,\operatorname{arsinh}{\left(a\tanh{\left(\tau\right)}\right)}.$$

It should clear $\mathcal{I}$ is equivalent to your integral up to an overall scale factor and a rescaling of the integration variable.

Note: The integral $\mathcal{I}{\left(a,z\right)}$ is odd in $a$ and even $z$, and the integral vanishes if either $a$ or $z$ is zero. So, it will suffice evaluate the integral for the $0<a\land0<z$ case.

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Suppose $\left(a,z\right)\in\mathbb{R}_{>0}^{2}$. Observing that

$$0<-a+\sqrt{1+a^{2}}<-a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}<1<a+\sqrt{1+a^{2}},$$

set

$$p:=a+\sqrt{1+a^{2}},$$

$$q:=-a+\sqrt{1+a^{2}},$$

$$w:=-a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}.$$

With the appropriate substitutions, we can rewrite the integral $\mathcal{I}$ as

$$\begin{align} \mathcal{I}{\left(a,z\right)} &=\int_{0}^{z}\mathrm{d}\tau\,\operatorname{arsinh}{\left(a\tanh{\left(\tau\right)}\right)}\\ &=\int_{0}^{\tanh{\left(z\right)}}\mathrm{d}y\,\frac{\operatorname{arsinh}{\left(ay\right)}}{1-y^{2}};~~~\small{\left[\tau=\operatorname{artanh}{\left(y\right)}\right]}\\ &=\int_{0}^{a\tanh{\left(z\right)}}\mathrm{d}x\,\frac{a\operatorname{arsinh}{\left(x\right)}}{a^{2}-x^{2}};~~~\small{\left[ay=x\right]}\\ &=\int_{0}^{a\tanh{\left(z\right)}}\mathrm{d}x\,\frac{a\ln{\left(x+\sqrt{1+x^{2}}\right)}}{a^{2}-x^{2}}\\ &=\int_{1}^{-a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}}\mathrm{d}t\,\frac{(-1)(1+t^{2})}{2t^{2}}\cdot\frac{a\ln{\left(\frac{1}{t}\right)}}{a^{2}-\left(\frac{1-t^{2}}{2t}\right)^{2}};~~~\small{\left[-x+\sqrt{1+x^{2}}=t\right]}\\ &=-\int_{-a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}}^{1}\mathrm{d}t\,\frac{2a\left(1+t^{2}\right)\ln{\left(t\right)}}{4a^{2}t^{2}-\left(1-t^{2}\right)^{2}}\\ &=-\int_{-a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}}^{1}\mathrm{d}t\,\frac{2a\left(1+t^{2}\right)\ln{\left(t\right)}}{\left[\left(a+\sqrt{1+a^{2}}\right)^{2}-t^{2}\right]\left[t^{2}-\left(-a+\sqrt{1+a^{2}}\right)^{2}\right]}\\ &=-\int_{w}^{1}\mathrm{d}t\,\frac{\left(p-q\right)\left(1+t^{2}\right)\ln{\left(t\right)}}{\left(p^{2}-t^{2}\right)\left(t^{2}-q^{2}\right)}.\\ \end{align}$$

Then, using the facts that $pq=1$ and $\frac{p^{2}+1}{p(p+q)}=\frac{1+q^{2}}{q(p+q)}=1$, we find

$$\begin{align} \mathcal{I}{\left(a,z\right)} &=-\int_{w}^{1}\mathrm{d}t\,\frac{\left(p-q\right)\left(1+t^{2}\right)\ln{\left(t\right)}}{\left(p^{2}-t^{2}\right)\left(t^{2}-q^{2}\right)}\\ &=-\frac{1}{p+q}\int_{w}^{1}\mathrm{d}t\,\left[\frac{p^{2}+1}{p^{2}-t^{2}}+\frac{1+q^{2}}{t^{2}-q^{2}}\right]\ln{\left(t\right)}\\ &=-\frac{1}{p+q}\int_{w}^{1}\mathrm{d}t\,\frac{p^{2}+1}{p^{2}-t^{2}}\ln{\left(t\right)}-\frac{1}{p+q}\int_{w}^{1}\mathrm{d}t\,\frac{1+q^{2}}{t^{2}-q^{2}}\ln{\left(t\right)}\\ &=-\int_{w}^{1}\mathrm{d}t\,\frac{p}{p^{2}-t^{2}}\ln{\left(t\right)}-\int_{w}^{1}\mathrm{d}t\,\frac{q}{t^{2}-q^{2}}\ln{\left(t\right)}\\ &=-\int_{\frac{w}{p}}^{\frac{1}{p}}\mathrm{d}u\,\frac{1}{1-u^{2}}\ln{\left(pu\right)};~~~\small{\left[t=pu\right]}\\ &~~~~~-\int_{\frac{w}{q}}^{\frac{1}{q}}\mathrm{d}u\,\frac{1}{u^{2}-1}\ln{\left(qu\right)};~~~\small{\left[t=qu\right]}\\ &=\int_{\frac{w}{p}}^{\frac{1}{p}}\mathrm{d}u\,\frac{\ln{\left(pu\right)}}{u^{2}-1}-\int_{q}^{\frac{q}{w}}\mathrm{d}v\,\frac{\ln{\left(\frac{v}{q}\right)}}{v^{2}-1};~~~\small{\left[u=\frac{1}{v}\right]}\\ &=\int_{qw}^{q}\mathrm{d}u\,\frac{\ln{\left(pu\right)}}{u^{2}-1}-\int_{q}^{\frac{q}{w}}\mathrm{d}u\,\frac{\ln{\left(pu\right)}}{u^{2}-1}.\\ \end{align}$$

The antiderivative we need comes from the following derivative:

$$\frac{d}{du}\left[\chi_{2}{\left(u\right)}-\ln{\left(pu\right)}\operatorname{artanh}{\left(u\right)}\right]=\frac{\ln{\left(pu\right)}}{u^{2}-1};~~~\small{u\in(0,1)\land p>0},$$

where the Legendre chi function $\chi_{2}$ may be given by the integral representation

$$\chi_{2}{\left(z\right)}:=\int_{0}^{z}\mathrm{d}y\,\frac{\operatorname{artanh}{\left(y\right)}}{y};~~~\small{z\in\left[-1,1\right]}.$$

Finally, we obtain

$$\begin{align} \mathcal{I}{\left(a,z\right)} &=\int_{qw}^{q}\mathrm{d}u\,\frac{\ln{\left(pu\right)}}{u^{2}-1}-\int_{q}^{\frac{q}{w}}\mathrm{d}u\,\frac{\ln{\left(pu\right)}}{u^{2}-1}\\ &=\left[\chi_{2}{\left(q\right)}-\ln{\left(pq\right)}\operatorname{artanh}{\left(q\right)}\right]-\left[\chi_{2}{\left(qw\right)}-\ln{\left(pqw\right)}\operatorname{artanh}{\left(qw\right)}\right]\\ &~~~~~-\left[\chi_{2}{\left(\frac{q}{w}\right)}-\ln{\left(p\frac{q}{w}\right)}\operatorname{artanh}{\left(\frac{q}{w}\right)}\right]+\left[\chi_{2}{\left(q\right)}-\ln{\left(pq\right)}\operatorname{artanh}{\left(q\right)}\right]\\ &=2\,\chi_{2}{\left(q\right)}-\chi_{2}{\left(qw\right)}-\chi_{2}{\left(\frac{q}{w}\right)}-\ln{\left(w\right)}\left[\operatorname{artanh}{\left(\frac{q}{w}\right)}-\operatorname{artanh}{\left(qw\right)}\right]\\ &=2\,\chi_{2}{\left(q\right)}-\chi_{2}{\left(qw\right)}-\chi_{2}{\left(\frac{q}{w}\right)}+\ln{\left(\frac{1}{w}\right)}\operatorname{artanh}{\left(\frac{q(1-w^{2})}{w(1-q^{2})}\right)}\\ &=2\,\chi_{2}{\left(q\right)}-\chi_{2}{\left(qw\right)}-\chi_{2}{\left(\frac{q}{w}\right)}\\ &~~~~~+\ln{\left(\frac{1}{-a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}}\right)}\operatorname{artanh}{\left(\tanh{\left(z\right)}\right)}\\ &=z\operatorname{arsinh}{\left(a\tanh{\left(z\right)}\right)}+2\,\chi_{2}{\left(-a+\sqrt{1+a^{2}}\right)}\\ &~~~~~-\chi_{2}{\left(\frac{a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}}{a+\sqrt{1+a^{2}}}\right)}\\ &~~~~~-\chi_{2}{\left(\frac{-a\tanh{\left(z\right)}+\sqrt{1+a^{2}\tanh^{2}{\left(z\right)}}}{a+\sqrt{1+a^{2}}}\right)}.\blacksquare\\ \end{align}$$

A very tidy final result, considering the nightmarish expression WolframAlpha returns if you ask it to do the integral for you, wouldn't you say? Cheers :)

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