How to figure out this definite integral

Question

How to calculate this definite integral? $$\int_0^1\frac{x\cos x}{\sqrt{1+x^2}}\mathrm{d}x$$ I've been thinking about this for a few days but still have no ideas.

Answer 1

I got an answer with Kampé de Fériet Function. \begin{aligned} &\int_0^1{\frac{x\cos x}{\sqrt{1+x^2}}}\text{d}x\\ &=\int_0^{\ln \left( 1+\sqrt{2} \right)}{\text{sinh}x\cos \left( \text{sinh}x \right)}\text{d}x \ \ \ \ \ (x\rightarrow \mathrm{sinh}x) \\ &=\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n \right) !}}\int_0^{\ln \left( 1+\sqrt{2} \right)}{\text{sinh}^{2n+1}x\text{d}x} \\ &=-\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}{\frac{\Gamma \left( n+1 \right)}{\Gamma \left( 2n+1 \right) \Gamma \left( n+\frac{3}{2} \right)}}\\ &\ \ \ -\sqrt{\frac{\pi}{2}}\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{\left( -1 \right) ^n2^k}{\Gamma \left( 2n+1 \right) \Gamma \left( \frac{3}{2}+k \right) \Gamma \left( \frac{3}{2}+n \right) \Gamma \left( \frac{3}{2}+k+n \right)}}}\ \ \ \ \ (\text{with the help of MMA}) \\ &=-\frac{\pi}{2}\mathbf{L}_{-1}\left( 1 \right) -\sqrt{\frac{\pi}{2}}\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{\left( -1 \right) ^n2^k}{\Gamma \left( 2n+1 \right) \Gamma \left( \frac{3}{2}+k \right) \Gamma \left( \frac{3}{2}+n \right) \Gamma \left( \frac{3}{2}+k+n \right)}}} \\ &=-\frac{\pi}{2}\mathbf{L}_{-1}\left( 1 \right) -\frac{4\sqrt{2}}{\pi}\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{\left( 1 \right) \!_n\left( 1 \right) \!_k}{\left( 1 \right) \!_n\left( \frac{1}{2} \right) \!_n\left( \frac{3}{2} \right) \!_n\left( \frac{3}{2} \right) \!_k\left( \frac{3}{2} \right) \!_{n+k}}}}\frac{\left( -\frac{1}{4} \right) \!^n}{n!}\frac{2^k}{k!} \end{aligned} where $\mathbf{L}_{\nu}\left( z \right)$ is modified Struve function and by the definition of Kampé de Fériet Function $$\displaystyle \mathbf{F}_{o,m,n}^{r,p,q}\left( \left. \begin{array}{c} \mathbf{A}_r;\mathbf{a }_p;\mathbf{a }_{q}^{'}\\[5pt] \mathbf{B}_o;\mathbf{b }_m;\mathbf{b }_{n}^{'}\\ \end{array} \right|x,y \right) =\sum_{i,j=0}^{\infty}{\frac{\prod_{s=1}^r{\left( A_s \right) \!_{i+j}}\prod_{k=1}^p{\left( a _k \right) \!_i}\prod_{\ell =1}^q{\left( a _{\ell}^{'} \right) \!_j}}{\prod_{v=1}^o{\left( B_v \right) \!_{i+j}}\prod_{t=1}^m{\left( b _t \right) \!_i}\prod_{u=1}^n{\left( b _{u}^{'} \right) \!_j}}}\frac{x^i}{i!}\frac{y^j}{j!}$$ where $\mathbf{a }_p=\alpha _1,\alpha _2,...,\alpha _p.$ Hence we have $$\int_0^1{\frac{x\cos x}{\sqrt{1+x^2}}}\text{d}x=-\frac{\pi}{2}\mathbf{L}_{-1}\left( 1 \right) -\frac{4\sqrt{2}}{\pi}\mathbf{F}_{1,3,1}^{0,1,1}\left( \left. \begin{array}{c} 1;1\\ \dfrac{3}{2};1,\dfrac{1}{2},\dfrac{3}{2};\dfrac{3}{2}\\ \end{array} \right|-\frac{1}{4},2 \right) $$