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As I need to study calculus for Physics in Grade 11, I learnt quite a bit and I can say I'm familiar with differentiation and Basic Integrals. But As I'm too curious for my own good, I decided to try out $\mathbf {Integration}$ $\mathbf{By}$ $\mathbf{Parts}$. I solved some easy ones. Then I thought, let's solve $$\int{x \sec x dx}$$

As I tried to solve it, I came up with this :

$$\int{x \sec x dx}$$
$$=x \int \sec x dx - \int \frac d{dx} (x) \left(\int \sec xdx\right)dx$$ $$=x \ln|\sec x + \tan x| - \int \ln |\sec x + \tan x| dx$$ $$=x \ln|\sec x + \tan x| - \ln|\sec x + \tan x| \int dx + \int \frac d{dx}(\ln|\sec x + \tan x|) (\int \ln|\sec x +\tan x|dx)dx$$ $$= x\ln |\sec x + \tan x| - x\ln|\sec x + \tan x| + \int \sec x (\int \ln |\sec x + \tan x| dx)dx$$

And it goes on, quite annoyingly.

I'm new to this, so I might have made some fundamental mistakes, but I still don't think this can be solved. A Mathematics book written in my vernacular informed me that some integral expressions cannot be simplified, so I guess this expression can be like it. Kindly look into this and let me know anything about it. Thanks in advance!

rikusp2002
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1 Answers1

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You'll need something called the polylogarithm function, $\operatorname{Li}_s(z):=\sum_{n\ge 1}z^nn^{-s}$. In particular $\frac{d}{dz}\operatorname{Li}_2(z)=-\frac{\ln(1-z)}{z}$ and$$\frac{d}{dx}(\operatorname{Li}_2(-ie^{ix})-\operatorname{Li}_2(ie^{ix}))=-i\ln\frac{1+ie^{ix}}{1-ie^{ix}},$$while$$\frac{d}{dx}\left(x\ln\frac{1-ie^{ix}}{1+ie^{ix}}\right)=\ln\frac{1-ie^{ix}}{1+ie^{ix}}+x\frac{2e^{ix}}{1+e^{2ix}}.$$The second term is $x\sec x$, so$$\int x\sec xdx=i(\operatorname{Li}_2(-ie^{ix})-\operatorname{Li}_2(ie^{ix}))+x\ln\frac{1-ie^{ix}}{1+ie^{ix}}+C.$$

J.G.
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