Evaluate ∫π/201+tanhx1+tanxdx \int_{0}^{\pi/2}\frac{1+\tanh x}{1+\tan x}dx

I need the method to evaluate this integral (the closed-form if possible).
π/201+tanhx1+tanxdx
I used the relationship between tanx and tanhx but it didn’t work. Any help?

Answer

I=π/201+tanh(x)1+tan(x)dx=π4+π/20tanh(x)1+tan(x)dx

I1=π/20tanh(x)1+tan(x)dx


tanh(x)=sinh(x)cosh(x)=exexex+ex=12exex+ex
so:
I1=π42π/2011+tan(x)exex+exdx
I2=π/2011+tan(x)exex+exdx


exex+ex=e2x1(e2x)=e2xn=0(1)ne2nx=n=0(1)ne2(n+1)x
and so:
I2=π/2011+tan(x)n=0(1)ne2(n+1)xdx=n=0(1)nπ/20e2(n+1)x1+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 π/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:
e2.5(n+1)xe2(n+1)x1+tan(x)e2.4(n+1)x
so if:
J(a)=π/20eaxdx=1eaπ/2a
so we have:
n=0(1)n1e2.5(n+1)π/22.5(n+1)I2n=0(1)n1e2.4(n+1)π/22.4(n+1)
and according to wolfram alpha these sums converge and arent too ugly, and we know that:
I=π/22I2
thats the best I can do at the moment I’ll take another look sometime. It is also worth noting that:
n=0(1)nn+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:
ln22.5+12.5n=11ne2.5nπ/2I2ln22.4+12.4n=11ne2.4nπ/2
we will try and focus on sums of the form:
S(α)=n=1exp(αn)n=ln(eα(eα1))
according to wolfram alpha, which we are able to simplify to:
S(α)=αln(eα1)
S(α)α=1ln(eα1)α

Attribution
Source : Link , Question Author : E.H.E , Answer Author : Henry Lee

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