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Through the formal substitutions $x=\arctan u$ and $u=v^{1/4}$ we get: $$I=\int_{0}^{\pi/4}\tan(2x)\,dx = \int_{0}^{\pi/4}\frac{2\tan(x)}{1-\tan^2(x)}\,dx = \int_{0}^{1}2u(1-u^4)^{-1}\,du \tag{1}$$ from which: $$I = \frac{1}{2}\int_{0}^{1}v^{-3/4}(1-u)^{-1}\,du = \frac{1}{2}\,B\left(\frac{1}{4},0\right)=\frac{\Gamma\left(\frac{1}{4}\right)\Gamma(0)}{2\,\... 2 A primitive of \tan(2x) is clearly -\frac{1}{2}\ln(\cos(2x)). This primitive is undefined for x=-\frac{\pi}{4}. Thus the integral is not defined. 1 Just an addendum, maybe some of this will be useful to you:$$\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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Here are some aspects to master such expressions: Let's denote the last expression with $G(x)$. \begin{align*} G(x)&=-A\frac14\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n-1} -\frac{B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n-1}\\ &\quad+\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(...

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The Bessel Function of the first kind and order $p$ has series representation given by $$J_p (x) = \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+p+1) } \left( \frac x 2 \right)^{2n+p}$$ For $p=1/2$, we find \begin{align} J_{1/2} (x) &= \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+3/2) } \left( \frac x 2 \right)^{2n+1/2} \\\\ &=\sqrt{\frac {... 0 As gammatester told the initial integral is related to Marcum Q-functionQ_M(\alpha,\beta)=\frac{1}{\alpha^{M-1}}\int_\beta^\infty x^Me^{-\frac{x^2+ \alpha^2}{2}}\mathrm{I}_{M-1}(\alpha x)dx$$in the following manner:$$\begin{eqnarray} \int_{0}^{b}xe^{-\,{x^{2} + z^{2} \over 2\sigma^2}} {\rm I}_{0}\left(\vphantom{\large A}xz \over \sigma^{2}\right)\,{\...

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For computing the principal values of such integrals, you may exploit the identities: $$\mathcal{L}(\text{ci}(x)) = -\frac{\log(1+s^2)}{s^2}\,\qquad \mathcal{L}(\text{si}(x))=\frac{1}{s}\arctan\left(\frac{1}{s}\right)\tag{1}$$ given by the Laplace transform and differentiation under the integral sign. For instance, since $$\mathcal{L}^{-1}\left(\frac{1}{q+... 4 New: In fact more or less the same argument shows that given E\subset\Bbb R there exists f:\Bbb R\to\Bbb R such that E is the set where f fails to have a limit if and only if E is an F_\sigma (a countable union of closed sets). First the answer to the OP, where E=\Bbb Q: Say g(x)=\sin(1/x) for x\ne0, g(0)=0. Let r_1,\dots be an ... 0 I think I've found a suitable [see EDIT below] way to generalize the fibonorial in the reals! (I've taken inspiration from an answer to this question.) The first terms of the fibonorial sequence \mathfrak{F}(n) are:$$\mathfrak{F}_7(n)=\color{red}{1},1,2,6,30,240,3120,65620$$where \mathfrak{F}_7(n) is just \mathfrak{F}(n) interrupted after the ... 3 If we set, using standard notations,$$ f(x) = -\frac{\pi x}{2}+\cos(x)+x\,\text{Si}(x) = x\,\text{si}(x)+\cos(x) \tag{1}$$we have:$$ f'(x) = -\frac{\pi}{2}+\text{Si}(x) = -\int_{x}^{+\infty}\frac{\sin t}{t}\,dt \tag{2}$$and since f(0)=1,$$ f(x) = 1-\int_{0}^{x}\int_{u}^{+\infty}\frac{\sin t}{t}\,dt\,du =1-\int_{1}^{+\infty}\frac{1-\cos(t x)}{t^2}\,dt\...

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Hint: Use integration by parts to show that $$-si(x)=\frac{\cos(x)}{x}-\int_x^{+\infty}\frac{\cos(t)}{t^2}dt=\frac{\cos(x)}{x}+\frac{\sin(x)}{x^2}-2\int_x^{+\infty}\frac{\sin(t)}{t^3}dt$$

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Extending what Zach466920 said: $$\partial_t[k(t,n)]=F(n+1)\cdot k(t,n+1)\tag1$$ We can develop a power series solution for this. Let $$k(t,n)=\sum_{m=0}^\infty \kappa(n, m)\ t^m$$ We equation $(1)$ as: $$\sum_{m=1}^\infty m\ \kappa(n, m)\ t^{m-1} =\sum_{m=0}^\infty(m+1)\ \kappa (n, m) \ t^m =F(n+1)\cdot\sum_{m=0}^\infty \kappa(n+1, m)\ t^m$$ Which gives ...

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$S(x)$ is an entire function, it does not have any logarithmic singularity: $$S(x)=\sum_{n\geq 1}\frac{x^n}{n\cdot n!}=\int_{0}^{x}\frac{e^t-1}{t}\,dt \tag{1}$$ also since $\frac{e^t-1}{t}$ is an entire function. The RHS of $(1)$ clearly depends on the exponential integral and clearly is not an elementary function.

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