I was trying to evaluate

$$\int \sec x dx \text{ using Complex Analysis }$$

I thought of rewriting it as $$\int \dfrac{1}{\Re(e^{\iota x})}$$ However, I cannot think of how to proceed from here. Could somebody please assist me in solving this Integral in the way I have tried? I would indeed be most grateful for your help. Many thanks!


Note: A similar method to evaluate $I=\int \cos x dx$:

$$\Re(e^{\iota x})=\cos x$$ $$\Longrightarrow \int \cos x dx = \Re \int e^{\iota x} dx$$ $$I=\Re(\dfrac{e^{\iota x}}{\iota}) = -\iota \Re(\cos x +\iota \sin x) = \sin x$$

  • $\begingroup$ I came across this method here $\endgroup$ – Ishan Jun 20 '15 at 8:33
  • $\begingroup$ Just to mention: I've never heard anybody refer to or intentionally write $i$ as $\iota$. If you want to do so, you can but I don't recommend it. $\endgroup$ – George V. Williams Jun 30 '15 at 16:12
  • $\begingroup$ I'm really sorry Sir. I had always thought that it was iota and not 'i'. I'll rectify it from now on. Thanks for telling me, Sir. $\endgroup$ – Ishan Jul 1 '15 at 9:10

A hint:

Subsitute the definition of $\cos$ to have:

$$ I = \int \frac{2}{e^{ix} + e^{-ix}} \, \mathrm{d}x = \frac{2}{i}\int \frac{i e^{ix}}{(e^{ix})^2 + 1} \, \mathrm{d}x = - 2 i \arctan e^{ix}$$

Then, you can use the definition of the inverse tangent when the argument is complex.

PS: Simplification should lead to a real result.

  • $\begingroup$ Thanks you Sir, but how would this use $\Im(e^{\iota x})$ or $\Re(e^{\iota x})$? $\endgroup$ – Ishan Jun 20 '15 at 20:18
  • $\begingroup$ Sir, could I ask for one last favour? Sir, could you please evaluate $$\int_0^{\pi/2} \dfrac{\sin^2(x)}{\sin (x) + \cos(x)} dx?$$ I keep getting my answer as $$\dfrac{\ln(\sqrt{2}+1)}{\sqrt{2}}$$However, my book lists a completely different answer. $\endgroup$ – Ishan Jun 20 '15 at 20:20
  • $\begingroup$ I think rewriting the integral as $\int \mathrm{d}x/\mathrm{Re}(e^{ix})$ may not lead you to a simple evaluation of the result. In general, it's not true that: $1/\mathrm{Re}(z) = \mathrm{Re}(1/z)$, for a complex number $z$ so you cannot use your approach anymore. $\endgroup$ – Dmoreno Jun 20 '15 at 20:27
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    $\begingroup$ Alright Sir. Thanks a lot for your help. Regarding the second problem I think I've understood where I've made a mistake although I still have to check. If I still get it wrong, I'll put itin a new post. $\endgroup$ – Ishan Jun 20 '15 at 20:32
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    $\begingroup$ Oh, sorry Sir, I forgot to click it. I just accepted it Sir. I'm really really sleepy and am now going to bed (it's past 2 at night here in India). Goodnight Sir.$$$$ PS. I just checked my answer again and it comes out to be correct. There's a misprint in my book due to which my answer had seemed to be wrong. $\endgroup$ – Ishan Jun 20 '15 at 20:37

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