$$\int_0^{\infty}\frac{dx}{2+\cosh (x)}$$

I'm not sure how to approach this problem. I tried substituting $\cosh (x) = \frac{e^x + e^{-x}}{2}$, and following the substitution $y = e^x$, I ended up with the integral

$$2 \cdot \int_1^{\infty} \frac{1}{(y+2)^2-3} dy$$

Which I wasn't able to evaluate, nor am I sure it is the best form to proceed.


  • $\begingroup$ There's a half-angle tanh substitution that works just like the hafl-angle tan substitution. en.wikipedia.org/wiki/… $\endgroup$ – B. Goddard Sep 21 '17 at 16:05
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    $\begingroup$ And I think you should be able to do your last integral. (I think the 4 should be a 3, however.) Let $u=y+2$ and do partial fractions (or use arctanh.) $\endgroup$ – B. Goddard Sep 21 '17 at 16:07
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    $\begingroup$ It is $$\int \frac{1}{(y+2)^2-3} dy$$ $\endgroup$ – Nosrati Sep 21 '17 at 16:10

$$\int_0^{\infty}\frac{dx}{2+\cosh (x)}=\int_0^{\infty}\frac{dx}{2+\frac{e^x+e^{-x}}{2}}$$ substitute $e^x=y\to x =\log y\to dx =\dfrac{dy}{y}$

limits become $x=0\to y=1;\;x\to\infty,\;y\to\infty$ so the integral becomes


$$=2\int_1^{\infty}\frac{dy}{(y+2+\sqrt 3)(y+2-\sqrt 3)}=\dfrac{1}{\sqrt 3}\int_1^{\infty}\left(\frac{1}{y+\sqrt{3}+2}-\frac{1}{y-\sqrt{3}+2}\right)\,dy=$$

$$=\frac{1}{\sqrt 3}\left(\lim_{M\to\infty}\left[\log(M+\sqrt{3}+2)-\log(M-\sqrt{3}+2)\right]-\log\dfrac{1}{3+\sqrt 3}+\log\dfrac{1}{3-\sqrt 3}\right)=$$

$$=\frac{1}{\sqrt 3}\left(\lim_{M\to\infty} \log\dfrac{M+\sqrt{3}+2}{M-\sqrt{3}+2}+\log\frac{3+\sqrt 3}{3-\sqrt 3}\right)=\frac{\log(2+\sqrt 3)}{\sqrt 3}$$

Hope this helps

  • $\begingroup$ That's what I did, too. I mistyped my question here. However, the correct answer is given as $$\frac{\sqrt2 \ln(\sqrt 2+1)}{2}$$ $\endgroup$ – John Lou Sep 21 '17 at 17:28
  • $\begingroup$ @JohnLou No way! The solution is $$\frac{2 \coth ^{-1}\left(\sqrt{3}\right)}{\sqrt{3}}=\frac{\log \left(2+\sqrt{3}\right)}{\sqrt{3}}$$ wolframalpha.com/input/… $\endgroup$ – Raffaele Sep 21 '17 at 17:51
  • $\begingroup$ Then it must be a mistake of the key. Thanks for the help. $\endgroup$ – John Lou Sep 21 '17 at 17:51

Using your substitution I get $$2\int_1^\infty\frac{dy}{y^2+4y+1}= 2\int_3^\infty\frac{du}{u^2-3}.$$ This can be done by partial fractions.


An alternative approach for interest. With this integral it's relatively easy to rewrite $\cosh$ in terms of e, but in cases where it's less feasible, you can use the half-angle substitution for the hyperbolic tangent.

Using that, we have, $$t = \tanh \frac x 2$$ $$\mathrm dx = \frac 2 {1-t^2} \mathrm dt$$ $$\cosh x = \frac{1+t^2}{1-t^2}$$

So for our limits we have,

$$x = 0 \implies t = \tanh(0) = 0$$ $$x \to \infty \implies t = \lim_{t_0 \to \infty} \tanh\left(t_0\right) = 1$$

And for our integrand we have,

$$\int_0^1 \frac {\mathrm dt} {2 + \frac {1+t^2}{1-t^2}} \cdot \frac 2 {1-t^2}$$ $$2\int_0^1 \frac {\mathrm dt} {2 - 2t^2 + 1 + t^2}$$ $$2\int_0^1 \frac {\mathrm dt} {3 - t^2}$$

Which is easily done with partial fractions, or another substitution. You will arrive at the same result as the others. If you want derivations of the above, I can edit this answer to include them.


write your Integrand in the form $$\frac{2e^x}{e^{2x}+4e^x+1}$$ and set $$e^x=t$$

  • $\begingroup$ I did do that, as I said in my question. $\endgroup$ – John Lou Sep 21 '17 at 17:30
  • $\begingroup$ and have you had success? $\endgroup$ – Dr. Sonnhard Graubner Sep 21 '17 at 17:36
  • $\begingroup$ I got an answer which is different from the answer in the key but the same as what was found by @raffaele. $\endgroup$ – John Lou Sep 21 '17 at 17:41
  • $\begingroup$ should is post you my answer? $\endgroup$ – Dr. Sonnhard Graubner Sep 21 '17 at 17:42
  • $\begingroup$ Were you able to get the answer in the form $\frac{\sqrt2 \ln ( \sqrt2 +1)}{2}$? $\endgroup$ – John Lou Sep 21 '17 at 17:50

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