1-sin-u-u-du-

Question Number 93061 by Ar Brandon last updated on 10/May/20

$$\int_{\mathrm{1}} ^{+\infty} \frac{\mathrm{sin}\:\mathrm{u}}{\mathrm{u}}\mathrm{du} \\ $$

Commented by mathmax by abdo last updated on 10/May/20

we have(π/2) =∫_0 ^∞ ((sinu)/u)du =∫_0 ^1 ((sinu)/u)du +∫_1 ^(+∞) ((sinu)/u)du ⇒ ∫_1 ^(+∞) ((sinu)/u)du =(π/2)−∫_0 ^1 ((sinu)/u)du but sinu =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))u^(2n+1) ⇒ ((sinu)/u) =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))u^(2n) ⇒∫_0 ^1 ((sinu)/u)du =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))×(1/(2n+1)) ⇒ ∫_1 ^(+∞) ((sinu)/u)du =(π/2)−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!(2n+1))) approximate value for n=4 ⇒ ∫_1 ^(+∞) ((sinu)/u)du∼(π/2)−(1−(1/(3.3!)) +(1/(5.5!))−(1/(7.7!)) +(1/(9.9!))) ⇒ ∫_1 ^(+∞) ((sinu)/u)du ∼(π/2)−1 +(1/(3.3!))−(1/(5.5!))+(1/(7.7!))−(1/(9.9!))

$${we}\:{have}\frac{\pi}{\mathrm{2}}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sinu}}{{u}}{du}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sinu}}{{u}}{du}\:+\int_{\mathrm{1}} ^{+\infty} \:\frac{{sinu}}{{u}}{du}\:\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{sinu}}{{u}}{du}\:=\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sinu}}{{u}}{du}\:\:{but}\:{sinu}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{u}^{\mathrm{2}{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\frac{{sinu}}{{u}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{u}^{\mathrm{2}{n}} \:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sinu}}{{u}}{du}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}×\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{sinu}}{{u}}{du}\:=\frac{\pi}{\mathrm{2}}−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$${approximate}\:{value}\:{for}\:{n}=\mathrm{4}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{sinu}}{{u}}{du}\sim\frac{\pi}{\mathrm{2}}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{3}!}\:+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{5}!}−\frac{\mathrm{1}}{\mathrm{7}.\mathrm{7}!}\:+\frac{\mathrm{1}}{\mathrm{9}.\mathrm{9}!}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{sinu}}{{u}}{du}\:\sim\frac{\pi}{\mathrm{2}}−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{5}.\mathrm{5}!}+\frac{\mathrm{1}}{\mathrm{7}.\mathrm{7}!}−\frac{\mathrm{1}}{\mathrm{9}.\mathrm{9}!} \\ $$

Commented by Ar Brandon last updated on 11/May/20

thank you

Commented by abdomathmax last updated on 14/May/20

$${you}\:{are}\:{welcome}. \\ $$

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