How to prove $\int_{0}^{\pi/4}\exp\left({-\sum_{n=1}^{\infty} {\frac {\tan^{2n}{x}}{n+\frac12}}}\right) dx =\ln\sqrt 2$? Question:-$$\int_{0}^{\pi/4}\exp\left({-\sum_{n=1}^{\infty} {\frac {\tan^{2n}{x}}{n+\frac12}}}\right) dx =\ln\sqrt 2$$
If we evaluate series separately then it diverges and it converges for $x\in[0,\frac{\pi}{4}]$,which are limits of integral.So,we have to consider it with integral. I don't know to evaluate series with the help of limits of integral.

Found another integral similar to first

Can anybody tell is it correct,if yes then how to Prove?
 A: Long Comment:
Some closed form results from Mathematica for reference, that maybe can be simplified further:
$$\int_0^{\frac{\pi }{4}} \exp \left(-\frac{1}{m}\sum _{n=1}^{\infty } \frac{\tan ^{2 n-1}(x)}{ \left(n-\frac{1}{2}\right)}\right) \, dx=\left(\frac{1}{4}+\frac{i}{4}\right) m \left(\, _2F_1\left(1,1;1+\frac{1}{m};\frac{1}{2}+\frac{i}{2}\right)-(1+i) \, _2F_1\left(1,\frac{1}{m};1+\frac{1}{m};i\right)\right)\tag{1}$$
$$\int_0^{\frac{\pi }{4}} \exp \left(-\frac{1}{m}\sum _{n=1}^{\infty } \frac{\tan ^{2 n}(x)}{ \,n}\right) \, dx=\frac{1}{2} \sqrt{\pi } \, _2\tilde{F}_1\left(\frac{1}{2},1;\frac{3}{2}+\frac{1}{m};-1\right) \Gamma \left(1+\frac{1}{m}\right)\tag{2}$$
both with $\Re\left(\frac{1}{m}\right)>-1$
$$\int_0^{\frac{\pi }{4}} \exp \left(-\frac{1}{m}\sum _{n=1}^{\infty } \frac{\tanh ^{2 n-1}(x)}{\left(n-\frac{1}{2}\right)}\right) \, dx=\frac{1}{2} \left(1-e^{-\frac{\pi }{2 m}}\right) m\tag{3}$$
$$\int_0^{\frac{\pi }{4}} \exp \left(-\frac{1}{m}\sum _{n=1}^{\infty } \frac{\tanh ^{2 n}(x)}{\, n}\right) \, dx=\frac{1}{2} \left(\frac{\sqrt{\pi }\, \Gamma \left(\frac{1}{m}\right)}{\Gamma \left(\frac{1}{2}+\frac{1}{m}\right)}-B_{\text{sech}^2\left(\frac{\pi }{4}\right)}\left(\frac{1}{m},\frac{1}{2}\right)\right)\tag{4}$$
with all four power series summations in $\tan(x)$ and $\tanh(x)$ equating to a $\log$ function.
$$\sum _{n=1}^{\infty } \frac{\tan ^{2 n-1}(x)}{n-\frac{1}{2}}=2 \tanh ^{-1}(\tan (x))=\log\left(\frac{1+\tan(x)}{1-\tan(x)} \right)$$
$$\sum _{n=1}^{\infty } \frac{\tan ^{2 n}(x)}{n}=-\log \left(1-\tan ^2(x)\right)$$
$$\sum _{n=1}^{\infty } \frac{\tanh ^{2 n-1}(x)}{n-\frac{1}{2}}=2 \tanh ^{-1}(\tanh (x))=\log\left(\frac{1+\tanh(x)}{1-\tanh(x)} \right)$$
$$\sum _{n=1}^{\infty } \frac{\tanh ^{2 n}(x)}{n}=-\log \left(\text{sech}^2(x)\right)$$
A: For the second integral (3.487), we have
\begin{align*}
&\int_0^{\pi /4} {\exp \left( { - \sum\limits_{n = 0}^\infty  {\frac{{\tan ^{2n + 1} x}}{{n + \frac{1}{2}}}} } \right)dx} = \int_0^{\pi /4} {\exp \left( { - 2\tanh ^{ - 1} (\tan x)} \right)dx} \\ & = \int_0^{\pi /4} {\frac{{1 - \tan x}}{{1 + \tan x}}dx}  = \int_0^1 {\frac{{1 - s}}{{1 + s}}\frac{1}{{1 + s^2 }}ds}  = \int_0^1 {\frac{1}{{1 + s}}ds}  - \int_0^1 {\frac{s}{{1 + s^2 }}ds} \\ & = \log 2 - \frac{1}{2}\log 2 = \frac{1}{2}\log 2.
\end{align*}
Half of what is expected.
A: As a start, according to Wolfy,
$\sum_{n=1}^{\infty} x^n/(n+1/2)
= -2(1-\tanh^{(-1)}(\sqrt{x})/\sqrt{x})$.
