Question Number 195895 by Erico last updated on 12/Aug/23
$$\mathrm{Calcul}\:\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \mathrm{t}\sqrt{\mathrm{tan}\left(\mathrm{t}\right)}\:\mathrm{dt} \\ $$
Answered by witcher3 last updated on 13/Aug/23
![∫_0 ^∞ ((tan^(−1) (t).(√t))/(1+t^2 )) =∫_0 ^∞ ((√t)/(1+t^2 )).∫_0 ^1 ((tdx)/(1+(xt)^2 )).dx =∫_0 ^1 ∫_0 ^∞ (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))dtdx ∫_0 ^∞ (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))dt=A ∫_(−∞) ^0 (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))=∫_0 ^∞ ((−it^(3/2) )/((1+t^2 )(1+x^2 t^2 )))dt=A ∫_C (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))=(((i)^(3/2) )/((2i)(1−x^2 ))).+(((i)^(3/2) x^2 )/(x^(3/2) (x^2 −1).2ix)) =(i^(1/2) /(2(1−x^2 )))(1−(1/( (√x)))) =((i^(3/2) π)/( (√x)(1+x)(1+(√x))))=((−πe^((i3π)/4) )/( (√x)(1+(√x))(1+x))). ∫_0 ^1 ((π(√2))/((1+t)(1+t^2 ))). =∫_0 ^1 ((π(√2)[1+t^2 +(1−t)(1+t))/(2(1+t)(1+t^2 ))) =(π/( (√2))){∫_0 ^1 (dt/(1+t))+∫_0 ^1 (dt/(1+t^2 ))−∫_0 ^1 ((tdt)/(1+t^2 ))} =(π/( (√2))){ln(2)+(π/4)−(1/2)ln(2))} =(π/( (√2)))[((ln(2))/2)+(π/4)]=∫_0 ^(π/2) x(√(tan (x)))dx](https://www.tinkutara.com/question/Q195957.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}\right).\sqrt{\mathrm{t}}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{\mathrm{t}}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }.\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tdx}}{\mathrm{1}+\left(\mathrm{xt}\right)^{\mathrm{2}} }.\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)}\mathrm{dtdx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt}=\mathrm{A} \\ $$$$\int_{−\infty} ^{\mathrm{0}} \frac{\mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)}=\int_{\mathrm{0}} ^{\infty} \frac{−\mathrm{it}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt}=\mathrm{A} \\ $$$$\int_{\mathrm{C}} \frac{\mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)}=\frac{\left(\mathrm{i}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\left(\mathrm{2i}\right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}.+\frac{\left(\mathrm{i}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right).\mathrm{2ix}} \\ $$$$=\frac{\mathrm{i}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right) \\ $$$$=\frac{\mathrm{i}^{\frac{\mathrm{3}}{\mathrm{2}}} \pi}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)}=\frac{−\pi\mathrm{e}^{\frac{\mathrm{i3}\pi}{\mathrm{4}}} }{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)\left(\mathrm{1}+\mathrm{x}\right)}. \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi\sqrt{\mathrm{2}}}{\left(\mathrm{1}+\mathrm{t}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi\sqrt{\mathrm{2}}\left[\mathrm{1}+\mathrm{t}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{t}\right)\left(\mathrm{1}+\mathrm{t}\right)\right.}{\mathrm{2}\left(\mathrm{1}+\mathrm{t}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tdt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right\} \\ $$$$\left.=\frac{\pi}{\:\sqrt{\mathrm{2}}}\left\{\mathrm{ln}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\right)\right\} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}}\left[\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right]=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{x}\sqrt{\mathrm{tan}\:\left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$ \\ $$
Commented by Rodier97 last updated on 17/Aug/23
$$\mathrm{Hi}…..\mathrm{i}'\mathrm{m}\:\mathrm{having}\:\mathrm{trouble}\:\mathrm{understanding}\:\mathrm{how}\:\mathrm{you}\:\mathrm{did} \\ $$$$\mathrm{the}\:\mathrm{second}\:\mathrm{line}…\mathrm{i}'\mathrm{m}\:\mathrm{still}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{notice}\:\mathrm{that}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{xt}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{but}\:\mathrm{where}\:\mathrm{does}\:''\mathrm{tdx}''\:\mathrm{come}\:\mathrm{from}\:\mathrm{in}\:\mathrm{your}\:\mathrm{term} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{tdx}}{\mathrm{1}+\left(\mathrm{xt}\right)^{\mathrm{2}} }\:\mathrm{dx}\:\:??? \\ $$$$\mathrm{please}\:! \\ $$
Commented by witcher3 last updated on 20/Aug/23
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tdx}}{\mathrm{1}+\left(\mathrm{xt}\right)^{\mathrm{2}} }=\mathrm{a};\mathrm{xt}=\mathrm{y}\Rightarrow\mathrm{dy}=\mathrm{tdx}\Rightarrow\mathrm{a}=\int_{\mathrm{0}} ^{\mathrm{t}} \frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}\right) \\ $$