Can someone please show me where I am going wrong? It seems there is a contradiction in the formula for the the following integral.

$$\int\frac {du}{\sqrt{u^2 + s^2}} = \log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert$$

Now, can't we rewrite the integral as follows using the substitution z = -u?

\begin{align} \int\frac {du}{\sqrt{u^2 + s^2}} &= \int\frac {du}{\sqrt{(-u)^2 + s^2}} = \int\frac {-dz}{\sqrt{z^2 + s^2}}\\ & = -\log \bigl\lvert z + \sqrt{z^2 + s^2}\bigr\rvert = -\log \bigl\lvert (-u + \sqrt{u^2 + s^2})\bigr\rvert \end{align} But, clearly $\;\log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert \ne -\log \bigl\lvert -u + \sqrt{u^2 + s^2} \bigr\rvert$

Also, the quantity inside $\log \bigl\lvert u + \sqrt{u^2 + s^2})\bigr\rvert$ is clearly non-negative for any choice of $u$ and $s$, so why do we need the absolute value of the quantity in our integral?


Both are not equal . But you'll get a term which will get added to the arbitrary constant and get cancelled in definite integral.

$$I=-\log|-u+\sqrt{u^2+s^2}|+c = \log\bigg\vert\frac{1}{-u+\sqrt{u^2+s^2}}\bigg\vert+c$$

$$I = \log\bigg\vert\frac{u+\sqrt{u^2+s^2}}{-u^2+u^2+s^2}\bigg\vert+c=\log\vert u+\sqrt{u^2+s^2}\vert-\log s^2+c = \log\vert u+\sqrt{u^2+s^2}\vert +k$$

where $k=c-\log s^2$

And this gets cancelled in a definite integral, so both are equivalent.


You have a faster way, using some hyperbolic trigonometry:

Set $t=\dfrac us$, so that $\mathrm du=s\,\mathrm dt$. The integral becomes, supposing $s>0\,$: $$\int\frac {\mathrm du}{\sqrt{u^2 + s^2}}=\int\frac {\not\! s\,\mathrm dt}{\not \!s\sqrt{ t^2 + 1}}=\int\frac {\mathrm dt}{\sqrt{ t^2 + 1}}=\operatorname{argsinh}t$$ and it is known that $$\operatorname{argsinh}t =\ln\bigl(t+\sqrt{t^2+1}\bigr). $$

  • $\begingroup$ Nice to see this notation with arg ! It remembers to me the old time. Cheers. $\endgroup$ – Claude Leibovici Aug 14 at 6:33
  • $\begingroup$ @ClaudeLeibovici: Actually, I should have written $\operatorname{argsh}$, but I didn't dare ;o) $\endgroup$ – Bernard Aug 14 at 8:35

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