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One may replace $t$ by $it$, getting $$e^{2itx+t^2}=\sum_{n=0}^\infty (-1)^ni^n\frac{t^n}{n!}H_n(x)$$ then one may take the real part: $$e^{t^2}\cos2xt=\sum_{n=0}^\infty \frac{t^{2n}}{(2n)!}H_{2n}(x)$$ where we have assumed that $t,x$ are real numbers.

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A primitive of $\tan(2x)$ is clearly $-\frac{1}{2}\ln(\cos(2x))$. This primitive is undefined for $x=-\frac{\pi}{4}$. Thus the integral is not defined.

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It is enough to compute asymptotics for $$S_n = \sum_{k=1}^{n}\sqrt{k} = \frac{2}{3}\,n^{3/2}+\frac{1}{2} n^{1/2}\color{red}{-\frac{\zeta\left(\frac{3}{2}\right)}{4\pi}}+O\left(\frac{1}{\sqrt{n}}\right)\tag{1}$$ and $$T_n = \sum_{k=0}^{n}\sqrt{k+\frac{1}{2}} = \frac{1}{\sqrt{2}}\left(S_{2n+1}-\sqrt{2}\,S_n\right)\tag{2}$$ by summation by parts to find the ...

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Put $$S_n=\sum_{k=1}^n (\frac{3}{2\sqrt{2}}\sqrt{2k}-\frac{1}{\sqrt{2}}\sqrt{2k+1}-\frac{1}{\sqrt{2}}\sqrt{2k-2})$$ and $\displaystyle T_n=\sum_{k=1}^n\sqrt{j}$. Using that $\displaystyle T_{2n+2}=\sqrt{2}T_n+\sqrt{2}\sqrt{n+1}+\sum_{j=1}^n \sqrt{2j+1}$, you get a formula for $S_n$ using $T_n$ and $T_{2n+2}$. Now using this answer Euler-Maclaurin Summation ...

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Just an addendum, maybe some of this will be useful to you: $$\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x=$$ $$=1.2577468869\dots= \gamma_{1}(1,0) - \gamma=\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma=$$ =\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}...

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