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Question Number 613 by 123456 last updated on 11/Feb/15
Ξ(a,b,c,d)=∫_0 ^1 ((sin^a (πx)cos^b (πx))/(x^c (1−x)^d ))dx Ξ(1,1,1,1)=?
$$\Xi\left({a},{b},{c},{d}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{sin}^{{a}} \left(\pi{x}\right)\mathrm{cos}^{{b}} \left(\pi{x}\right)}{{x}^{{c}} \left(\mathrm{1}−{x}\right)^{{d}} }{dx} \\ $$$$\Xi\left(\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1}\right)=? \\ $$
Commented by prakash jain last updated on 10/Feb/15
Initial simplificaion Ξ(1,1,1,1)=∫_0 ^1 ((sin πx cos πx)/(x(1−x))) Ξ(a,b,0,0)=∫_0 ^1 sin^a (πx)cos^b (πx) dx Ξ(0,0,1−c,1−d)=∫_0 ^1 ((x^c (1−x)^d )/(x(1−x)))sin πxcos πxdx
$$\mathrm{Initial}\:\mathrm{simplificaion} \\ $$$$\Xi\left(\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{sin}\:\pi{x}\:\mathrm{cos}\:\pi{x}}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$\Xi\left({a},{b},\mathrm{0},\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{sin}^{{a}} \left(\pi{x}\right)\mathrm{cos}^{{b}} \left(\pi{x}\right)\:{dx} \\ $$$$\Xi\left(\mathrm{0},\mathrm{0},\mathrm{1}−{c},\mathrm{1}−{d}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{c}} \left(\mathrm{1}−{x}\right)^{{d}} }{{x}\left(\mathrm{1}−{x}\right)}\mathrm{sin}\:\pi{x}\mathrm{cos}\:\pi{xdx} \\ $$
Commented by prakash jain last updated on 11/Feb/15
∫_0 ^1 ((sin (πx)cos (πx))/(x(1−x)))=∫_0 ^1 ((sin 2πx)/(x(1−x))) x=0.5−u⇒dx=−du, x=0, u=0.5, x=1, u=−0.5 =−∫_(.5) ^(0.5) ((sin (π−2πu))/((0.5−u)(0.5+u))) =∫_(−.5) ^(0.5) ((sin 2πu)/((0.5)^2 −u^2 )) dx=0 (∵ odd function of u)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}\:\left(\pi{x}\right)\mathrm{cos}\:\left(\pi{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}\:\mathrm{2}\pi{x}}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$${x}=\mathrm{0}.\mathrm{5}−{u}\Rightarrow{dx}=−{du},\:{x}=\mathrm{0},\:{u}=\mathrm{0}.\mathrm{5},\:{x}=\mathrm{1},\:{u}=−\mathrm{0}.\mathrm{5} \\ $$$$=−\int_{.\mathrm{5}} ^{\mathrm{0}.\mathrm{5}} \frac{\mathrm{sin}\:\left(\pi−\mathrm{2}\pi{u}\right)}{\left(\mathrm{0}.\mathrm{5}−{u}\right)\left(\mathrm{0}.\mathrm{5}+{u}\right)} \\ $$$$=\int_{−.\mathrm{5}} ^{\mathrm{0}.\mathrm{5}} \frac{\mathrm{sin}\:\mathrm{2}\pi{u}}{\left(\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} −{u}^{\mathrm{2}} }\:{dx}=\mathrm{0}\:\left(\because\:\mathrm{odd}\:\mathrm{function}\:\mathrm{of}\:{u}\right) \\ $$

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