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Question Number 123898 by mnjuly1970 last updated on 29/Nov/20

$$...nicecalculus...$$$$findaseriesrepresentation$$$$forthefollowingintegral::$$$$\varphi ={\int }_0^{1}cos(h\left(xln\left(x\right)\right)dx$$$$$$

Answered by mindispower last updated on 29/Nov/20

$${\int }_0^1{ln}^a\left(x\right)dx={\int }_0^{\mathrm{\infty }}{(-t)}^a{e}^{-t}dt...1$$$$={(-1)}^a\mathrm{\Gamma }(a+1).....nice$$$$cos\left(x\right)=\sum _{n⩾0}\frac{{(-1)}^n{x}^{2n}}{\left(2n\right)!}$$$$\mathrm{\varnothing }={\int }_0^1\sum _{n⩾0}\frac{{(-1)}^n}{2n!}.{\left(hln\left(x\right)\right)}^{2n}dx$$$$=\sum _{n⩾0}\frac{{(-1)}^n}{\left(2n\right)!}{h}^{2n}{\int }_0^1{ln}^{2n}\left(x\right)dx....appie1a=2n$$$$=\sum _{n⩾0}\frac{{(-1)}^n}{\left(2n\right)!}{h}^{2n}.\frac{{(-1)}^{2n}\mathrm{\Gamma }(2n+1)}{}$$$$=\sum _{n⩾0}{(-h^2)}^n=\frac{1}{1+h^2}$$

Commented by Mammadli last updated on 29/Nov/20

Commented by Dwaipayan Shikari last updated on 29/Nov/20

$$\underset{x\to 1}{\lim }\frac{a(a^{x-1}-1)}{x-1}=a.log\left(a\right)$$

Commented by mnjuly1970 last updated on 29/Nov/20

$$thanksforyourwork$$$$sirpower.but$$$$h:hypbolic$$

Answered by Dwaipayan Shikari last updated on 29/Nov/20

$${\int }_0^{1}cos(h\left(log\left(x\right)\right)dx$$$$={\int }_0^1\frac{e^{logx}+e^{-logx}}{2}dx$$$$=\frac{1}{2}{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(logx\right)}^n}{n!}+\frac{1}{2}{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\left(logx\right)}^n}{n!}$$$$=\frac{1}{2}(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}{\int }_0^1{\left(logx\right)}^n+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}{\int }_0^1{\left(logx\right)}^n)$$$$=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}{\int }_{-\mathrm{\infty }}^0{e}^t{t}^n+\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}{\int }_{-\mathrm{\infty }}^0{e}^t{t}^{n}dtt=-\mathrm{\Phi }$$$$=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}{\int }_0^{\mathrm{\infty }}{e}^{-\mathrm{\Phi }}{\mathrm{\Phi }}^{n}d\mathrm{\Phi }+\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}{\int }_0^{\mathrm{\infty }}{e}^{-\mathrm{\Phi }}{\mathrm{\Phi }}^{n}d\mathrm{\Phi }$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{\mathrm{\Gamma }(n+1)}{2n!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+1)}{2n!}$$$$$$

Commented by mnjuly1970 last updated on 29/Nov/20

$$thankyousomuchsir$$$$payan.$$$$questionwascorrected:$$$$integrand:cos\left(h\left(xln\left(x\right)\right)\right)$$$$$$

Answered by Dwaipayan Shikari last updated on 29/Nov/20

$${\int }_0^{1}cos\left(h\left(xlogx\right)\right)dx$$$$={\int }_0^1\frac{e^{xlogx}+e^{-xlogx}}{2}dx$$$$=\frac{1}{2}\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{{\left(xlogx\right)}^n}{n!}-\frac{1}{2}\sum _{n⩾0}{(-1)}^n{\int }_0^1\frac{{\left(xlogx\right)}^n}{n!}(t(n+1)=-\mathrm{\Phi })$$$$=\frac{1}{2}\sum _{n⩾0}\frac{1}{n!}{\int }_{-\mathrm{\infty }}^0{e}^{(n+1)t}{t}^{n}dt-\frac{1}{2}\sum _{n⩾0}{(-1)}^n\frac{1}{n!}{\int }_{-\mathrm{\infty }}^0{e}^{t(n+1)}{t}^{n}dt$$$$=\frac{1}{2}\sum _{n⩾0}\frac{1}{n!}\frac{(-1)}{{(n+1)}^{n+1}}{\int }_0^{\mathrm{\infty }}{e}^{-\mathrm{\Phi }}{\mathrm{\Phi }}^{n}d\mathrm{\Phi }-\frac{1}{2}\sum _{n⩾0}\frac{{(-1)}^{2n}}{n!{(n+1)}^{n+1}}{\int }_0^{\mathrm{\infty }}{e}^{-\mathrm{\Phi }}{\mathrm{\Phi }}^{n}d\mathrm{\Phi }$$$$=\frac{1}{2}\sum _{n⩾0}\frac{{(-1)}^n}{{(n+1)}^{n+1}}-\frac{1}{2}\sum _{n⩾0}\frac{1}{{(n+1)}^{(n+1)}}$$$$=\frac{1}{2}(\frac{1}{1^1}-\frac{1}{2^2}+\frac{1}{3^3}-\frac{1}{4^4}+...-\frac{1}{1^1}-\frac{1}{2^2}-\frac{1}{3^3}-\frac{1}{4^4}+-..)$$$$=-(\frac{1}{2^2}+\frac{1}{4^4}+\frac{1}{6^6}+\frac{1}{8^8}+....)=-\sum _{n⩾1}\frac{1}{{\left(2n\right)}^{2n}}$$

Commented by mnjuly1970 last updated on 29/Nov/20

$$peacebeuponyou$$$$sirpayan$$$$tireless,powerful$$$$andunassuming.$$$$gratefulandgodkeepyou..$$

Commented by Dwaipayan Shikari last updated on 29/Nov/20

$$Withpleasuresir!youarereallygreatman$$

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