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Question Number 173139 by mnjuly1970 last updated on 07/Jul/22

Answered by Mathspace last updated on 07/Jul/22

$$\mathrm{{\rm Y}}={\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{1+x}(x^2+2x+2)}$$$$changement\sqrt{1+x}=tgive$$$$x=t^2-1\Rightarrow$$$$\mathrm{{\rm Y}}={\int }_1^{\mathrm{\infty }}\frac{2tdt}{t({(t^2-1)}^2+2(t^2-1)+2)}$$$$=2{\int }_1^{+\mathrm{\infty }}\frac{dt}{t^4-2t^2+1+2t^2-2+2}$$$$=2{\int }_1^{+\mathrm{\infty }}\frac{dt}{t^4+1}$$$$wehave{\int }_0^{\mathrm{\infty }}\frac{dt}{t^4+1}$$$$={\int }_0^1\frac{dt}{t^4+1}+{\int }_1^{\mathrm{\infty }}\frac{dt}{t^4+1}\Rightarrow$$$${\int }_1^{\mathrm{\infty }}\frac{dt}{t^4+1}={\int }_0^{\mathrm{\infty }}\frac{dt}{t^4+1}-{\int }_0^1\frac{dt}{t^4+1}$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{t^4+1}{=}_{t^4=z}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{z^{\frac{1}{4}-1}}{1+z}dz$$$$=\frac{1}{4}\times \frac{\pi }{sin\left(\frac{\pi }{4}\right)}=\frac{\pi }{4.\frac{1}{\sqrt{2}}}=\frac{\pi \sqrt{2}}{4}$$$${\int }_0^1\frac{dt}{1+t^4}={\int }_0^1\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{4n}dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^1{t}^{4n}dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{1}{4n+1}$$$$=\sum _{p=0}^{\mathrm{\infty }}\frac{1}{8p+1}-\sum _{p=0}^{\mathrm{\infty }}\frac{1}{4(2p+1)+1}$$$$=\sum _{p=0}^{\mathrm{\infty }}\frac{1}{8p+1}-\sum _{p=0}^{\mathrm{\infty }}\frac{1}{8p+5}$$$$=\frac{1}{8}\sum _{p=0}^{\mathrm{\infty }}(\frac{1}{p+\frac{1}{8}}-\frac{1}{p+\frac{5}{8}})$$$$=$$$$\frac{1}{16}\sum _{p=0}^{\mathrm{\infty }}\frac{1}{(p+\frac{1}{8})(p+\frac{5}{8})}$$$$=\frac{1}{16}\times \frac{\mathrm{\Psi }\left(\frac{5}{8}\right)-\mathrm{\Psi }\left(\frac{1}{8}\right)}{\frac{5}{8}-\frac{1}{8}}$$$$=\frac{1}{8}(\mathrm{\Psi }\left(\frac{5}{8}\right)-\mathrm{\Psi }\left(\frac{1}{8}\right))...$$$$$$

Commented by mnjuly1970 last updated on 07/Jul/22

$$\mathrm{bravo}\mathrm{sir}...$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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