Evaluate the integral $\int\limits_{0}^{b}\frac{dx}{\sqrt{a^2+x^2}}$

Question

Show that $$\int\limits_{0}^{b}\frac{dx}{\sqrt{a^2+x^2}}=\sinh^{-1}\frac{b}{a}$$

However, when I use Maple or WolframAlpha to calculate the left integration, both gave me $-\frac{\ln(a^2)}{2}+\ln(b+\sqrt{a^2+b^2})$, which seems not agree with the result on the right.

[This integration is from the book Introduction to Superconductivity Second Edition By Michael Tinkham Page 56]

A snapshot of the textbook

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The $\Delta$ is independent of $\xi$.

This is from the famous BCS theory and I think it should be correct. Also, I have checked another book, which shows the same result ($\sinh^{-1}\frac{b}{a}$).

Answer 1

There is a link between $ sinh^{-1} $ and $ \ln $ via:

$$\sinh(x)=\frac{e^x-e^{-x}}{2}=y$$ $$x=\sinh^{-1}(y)$$ By solving for $ e^x $, the equation $$e^{2x}-2ye^x-1=0$$ we find $$e^x=y+\sqrt{y^2+1}$$ and then $$x=\sinh^{-1}(y)=\ln(y+\sqrt{y^2+1})$$

So if $a>0$, $$\sinh^{-1}\Bigl(\frac ba\Bigr)=\ln\Bigl(\frac ba+\sqrt{\frac{b^2}{a^2}+1}\Bigr)$$

$$=\ln\Bigl(\frac{b+\sqrt{a^2+b^2}}{a}\Bigr)$$ $$=\ln(b+\sqrt{b^2+a^2})-\ln(a)$$

Answer 2

They are indeed equivalent expressions. $${\text{sinh}}^{-1}(x)=\ln{\left(x+\sqrt{x^2+1}\right)}$$ For your problem, $x=\frac{b}{a}$: $${\text{sinh}}^{-1}\left(\frac{b}{a}\right)=\ln{\left(\frac{b}{a}+\sqrt{{\left(\frac{b}{a}\right)}^2+1}\right)}$$ $$=\ln{\left(b+\sqrt{b^2+a^2}\right)}-\ln{a}$$ $${\text{sinh}}^{-1}\left(\frac{b}{a}\right)=-\frac{\ln{(a^2)}}{2}+\ln{\left(b+\sqrt{a^2+b^2}\right)}$$