3

Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.584$ gives:

$$\int \dfrac {\mathrm d x} {p^2 + q^2 \cosh^2 a x} = \begin{cases} \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C \\ \dfrac 1 {a p \sqrt {p^2 + q^2} } \arctan \left({\dfrac {p \tanh a x} {\sqrt {p^2 + q^2} } }\right) + C \end{cases}$$

(As usual he glosses over negative argument to the $\ln$, but all you need to do is modulus it.)

He does not give the conditions under which either case holds. This is crucial.

First I multiplied top and bottom by $\operatorname {csch}^2 a x$, then used $\operatorname {csch}^2 a x = \coth^2 a x - 1$ for the bottom, then substituted $u = \coth a x$ to convert it into the form:

$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\frac {p^2} {p^2 + q^2} - u^2}$$

I can't see there is any other case than that $\dfrac {p^2} {p^2 + q^2}$ is positive, hence:

$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\left({\frac p {\sqrt {p^2 + q^2} } }\right)^2 - u^2}$$

This is a standard integral, giving:

$$\frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left|{\frac {\sqrt {p^2 + q^2} u - p} {\sqrt {p^2 + q^2} u + p} }\right| + C$$

which leads to:

$$\dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C$$

And that all works fine, no worries.

BUT, as far as I can see, there is no way to get to the arctangent result. To get there, we need $u^2 + k^2$ for some constant $k$, and I can't see how to get there from the given integrand.

azif00
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Prime Mover
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  • Are you sure it shouldn't be $\text{artanh } $ in stead of $\arctan$? – A-Level Student Dec 31 '20 at 22:43
  • @A-LevelStudent Good question, but I'm absolutely quite sure. If that were the case then the question would be solved as the arctanh construction is equivalent to the log of that quotient, as far as I remember without looking up which one goes on top and which one on the bottom. Besides, if that were the case, it would not be presented as a "cases" construct, but as an alternative with an "equals" separating the two results. (It actually says $\tan^{-1}$ but I dislike that. – Prime Mover Dec 31 '20 at 22:47
  • That's why I thought it may be $\text{artanh}$ as then it fits very nicely, but you have a point about there being 2 options. I am differentiating the $\arctan$ one now and seeing if it does work. – A-Level Student Dec 31 '20 at 22:52
  • There is a version of the sneaky Weierstrass substitution that applies to hyperbolic functions, – Will Jagy Dec 31 '20 at 23:01
  • @WillJagy but isn't that still in terms of $\tanh$ and not $\tan$? – A-Level Student Dec 31 '20 at 23:02
  • @A-LevelStudent neither one by nature. All it does is make a rational function, namely a polynomial divided by a polynomial. Then partial fractions, which is not entirely predictable – Will Jagy Dec 31 '20 at 23:05
  • @WillJagy thanks, I meant a substitution involving $\tanh$ as opposed to $\tan$. Good luck in solving it with that, I look forward to your results. – A-Level Student Dec 31 '20 at 23:08
  • @WillJagy Yes, I haven't tried that approach - although I'm aware of it, I usually find it's quicker when in hyperbolics to make the substitution $u = e^{ax}$ as this usually gets me where I want to go quicker than when i use the full $\tanh (ax/2)$ subst. I'll give it a go tomorrow perhaps when I'm fresh. – Prime Mover Dec 31 '20 at 23:13

2 Answers2

2

Note $$ \frac12\ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } } = \frac12\ln {\frac {\frac {p \tanh a x }{\sqrt {p^2 + q^2} } + 1}{\frac{p \tanh a x }{\sqrt {p^2 + q^2} }-1}} =\tanh^{-1}\frac {p \tanh a x }{\sqrt {p^2 + q^2} }\\ $$

So, there is a typo in the second expression.

Quanto
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1

The second result is wrong it would seem, at least if it's supposed to hold for all values of $p, q,a$. Letting $p=q=a=1$ and using my calculator to evaluate the integral between $0$ and $1$:

$$\int_0^1\frac{1}{1+\cosh^2 x}dx=0.4257352463...$$ But $$\frac{1}{\sqrt2}\arctan\frac{\tanh 1}{\sqrt2}-\frac{1}{\sqrt2}\arctan\frac{\tanh 0}{\sqrt2}=0.3493059377...$$

A-Level Student
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