Evaluate $ \int_{0}^{\pi/2}\frac{1+\tanh x}{1+\tan x}dx $

Solution 1:

$$I=\int_0^{\pi/2}\frac{1+\tanh(x)}{1+\tan(x)}dx=\frac\pi 4+\int_0^{\pi/2}\frac{\tanh(x)}{1+\tan(x)}dx$$

$$I_1=\int_0^{\pi/2}\frac{\tanh(x)}{1+\tan(x)}dx$$

$$\tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{e^x+e^{-x}}=1-2\frac{e^{-x}}{e^x+e^{-x}}$$ so: $$I_1=\frac{\pi}{4}-2\int_0^{\pi/2}\frac{1}{1+\tan(x)}\frac{e^{-x}}{e^x+e^{-x}}dx$$ $$I_2=\int_0^{\pi/2}\frac{1}{1+\tan(x)}\frac{e^{-x}}{e^x+e^{-x}}dx$$

$$\frac{e^{-x}}{e^x+e^{-x}}=\frac{e^{-2x}}{1-(-e^{-2x})}=e^{-2x}\sum_{n=0}^\infty(-1)^ne^{-2nx}=\sum_{n=0}^\infty(-1)^ne^{-2(n+1)x}$$ and so: $$I_2=\int_0^{\pi/2}\frac{1}{1+\tan(x)}\sum_{n=0}^\infty(-1)^ne^{-2(n+1)x}dx=\sum_{n=0}^\infty(-1)^n\int_0^{\pi/2}\frac{e^{-2(n+1)x}}{1+\tan(x)}dx$$ as for this integral its quite messy and I'm not sure what to do from here, It would be easier for $0$ to $\pi/4$ I think. I will say that as $n$ increases the terms get smaller very quickly so an approximation of the first few would be quite accurate if possible.

One possible way I have noticed is that: $$e^{-2.5(n+1)x}\le\frac{e^{-2(n+1)x}}{1+\tan(x)}\le e^{-2.4(n+1)x}$$ so if: $$J(a)=\int_0^{\pi/2}e^{-ax}dx=\frac{1-e^{-a\pi/2}}{a}$$ so we have: $$\sum_{n=0}^\infty(-1)^n\frac{1-e^{-2.5(n+1)\pi/2}}{2.5(n+1)}\le I_2\le \sum_{n=0}^\infty(-1)^n\frac{1-e^{-2.4(n+1)\pi/2}}{2.4(n+1)}$$ and according to wolfram alpha these sums converge and arent too ugly, and we know that: $$I=\pi/2-2I_2$$ thats the best I can do at the moment I'll take another look sometime. It is also worth noting that: $$\sum_{n=0}^\infty\frac{(-1)^n}{n+1}=\ln(2)$$ and the second part of the sum could be expanded into a double summation

Back to add a little to this answer, so far we know: $$\frac{\ln2}{2.5}+\frac 1 {2.5}\sum_{n=1}^\infty \frac 1ne^{-2.5n\pi/2}\le I_2\le \frac{\ln2}{2.4}+\frac 1 {2.4}\sum_{n=1}^\infty \frac 1ne^{-2.4n\pi/2}$$ we will try and focus on sums of the form: $$S(\alpha)=\sum_{n=1}^\infty\frac{\exp(-\alpha n)}{n}=-\ln(e^{-\alpha}(e^\alpha-1))$$ according to wolfram alpha, which we are able to simplify to: $$S(\alpha)=\alpha-\ln(e^\alpha-1)$$ $$\frac{S(\alpha)}{\alpha}=1-\frac{\ln(e^\alpha-1)}{\alpha}$$