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Evaluating this integral: ‍‍‍‍‍‍‍ $$\displaystyle \int \dfrac {\mathrm d x} {p^2 - q^2 \sinh^2 a x}$$

Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.556$ gives:

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

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

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

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

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:

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

As can be seen, the fraction is upside down in the argument of the $\ln$. This means there's a negative sign gone astray somewhere. I can't see where.

dua
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Prime Mover
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1 Answers1

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Wow you gave me such a hard time on this, I have never used these functions before but the mistake was simpler than I thought. This is just a simple sign mistake in the first step of your calculation.

$$\int \dfrac{\mathrm{sech}^2(ax) \mathrm{d}x}{\mathrm{sech}^2(ax) (p^2-q^2 \sinh^2(ax))} = \int \dfrac{\mathrm{sech}^2(ax) \mathrm{d}x}{p^2\mathrm{sech}^2(ax)-q^2 \tanh^2(ax)}$$

$$=\int \dfrac{(1-\tanh^2(ax)) \mathrm{d}x}{p^2(1-\tanh^2(ax))-q^2 \tanh^2(ax)}$$

$$= \int \dfrac{(1/a) \tanh'(ax) \mathrm{d}x}{p^2-(p^2+q^2) \tanh^2(ax)}$$

$$= \dfrac{1}{a(p^2+q^2)} \int \dfrac{ \tanh'(ax) \mathrm{d}x}{\left(\dfrac{p}{\sqrt{p^2+q^2}}\right)^2- \tanh^2(ax)} = - \dfrac{1}{a(p^2+q^2)}\int \dfrac{\mathrm{d}u}{u^2-\left(\dfrac{p}{\sqrt{p^2+q^2}}\right)^2}$$

Axel
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