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Question Number 136868 by mnjuly1970 last updated on 27/Mar/21

...... nice calculus.... prove that :: i: ∫_(−∞) ^( ∞) (dx/((1+x+e^x )^2 +π^2 )) =(2/3) ii: ∫_0 ^( ∞) ((sin(tan(x)))/x)dx=(π/2)(1−(1/e))

$$\:\:\:\:\:\:\:\:\:......\:{nice}\:\:\:\:\:{calculus}.... \\ $$$$\:\:{prove}\:\:{that}\::: \\ $$$$\:{i}:\:\:\int_{−\infty} ^{\:\infty} \frac{{dx}}{\left(\mathrm{1}+{x}+{e}^{{x}} \right)^{\mathrm{2}} +\pi^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:{ii}:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({tan}\left({x}\right)\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right) \\ $$$$\:\:\:\:\: \\ $$

Answered by Ñï= last updated on 27/Mar/21

I=∫_0 ^∞ ((sin (tan x))/x)dx=Σ_(n=0) ^∞ ∫_(n(π/2)) ^((n+1)(π/2)) ((sin (tan x))/x)dx ???? =Σ_(n=0) ^∞ ∫_(nπ) ^((2n+1)(π/2)) ((sin (tan x))/x)dx+Σ_(n=0) ^∞ ∫_((2n−1)(π/2)) ^(nπ) ((sin (tan x))/x)dx ???? =Σ_(n=0) ^∞ ∫_0 ^(π/2) ((sin (tan x))/(nπ+x))dx−Σ_(n=0) ^∞ ∫_0 ^(π/2) ((sin (tan x))/(nπ−x))dx =∫_0 ^(π/2) Σ_(n=0) ^∞ ((2x)/(x^2 −n^2 π^2 ))sin (tan x)dx =∫_0 ^(π/2) cot xsin (tan x)dx =∫_0 ^∞ ((sin u)/(u(1+u^2 )))du =∫_0 ^1 da∫_0 ^∞ ((cos (au))/(1+u^2 ))du =(π/2)∫_0 ^1 e^(−a) da =(π/2)(1−e^(−1) )

$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{{x}}{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{{n}\frac{\pi}{\mathrm{2}}} ^{\left({n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{{x}}{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:???? \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{{n}\pi} ^{\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{{x}}{dx}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\left(\mathrm{2}{n}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}} ^{{n}\pi} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{{x}}{dx}\:\:\:\:\:\:\:???? \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{{n}\pi+{x}}{dx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{{n}\pi−{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cot}\:{x}\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{u}}{{u}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {da}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\left({au}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{a}} {da} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{e}^{−\mathrm{1}} \right) \\ $$

Commented by mnjuly1970 last updated on 27/Mar/21

very nice .. thank you so much ...

$$\:\:{very}\:{nice}\:.. \\ $$$$\:\:\:{thank}\:{you}\:{so}\:{much}\:... \\ $$

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