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Question Number 199921 by Mingma last updated on 11/Nov/23

Answered by des_ last updated on 12/Nov/23

∫_0 ^∞ ((cos x)/(x^2 +1)) dx = I; I(a) = ∫_0 ^∞ ((cos(ax))/(x^2 +1)) dx, a >0; I(a→+0) = ∫_0 ^∞ (1/(x^2 +1)) dx = (π/2); (1) (dI/da) = ∫_0 ^∞ ((−xsin(ax))/(x^2 +1)) dx = −∫_0 ^∞ ((sin(ax))/x) dx + ∫_0 ^∞ ((sin(ax))/(x(x^2 +1))) dx ; (dI/da) = −(π/2) + ∫_0 ^∞ ((sin(ax))/(x(x^2 +1))) dx; (dI/da)(a→+0) = −(π/2); (2) (d^2 I/da^2 ) = ∫_0 ^∞ ((cos(ax))/(x^2 +1)) dx = I; (d^2 I/da^2 ) − I = 0 ⇒ I(a) = C_1 e^a + C_2 e^(−a) ; (1),(2) ⇒ { ((C_1 + C_2 = (π/2))),((C_1 − C_2 = −(π/2))) :} ⇒ { ((C_1 = 0)),((C_(2 ) = (π/2))) :} ; I(a) = (π/2) e^(−a) ; I = I(1) = (π/(2e))

$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:{I}; \\ $$$${I}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left({ax}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx},\:{a}\:>\mathrm{0}; \\ $$$$\:{I}\left({a}\rightarrow+\mathrm{0}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:\frac{\pi}{\mathrm{2}};\:\left(\mathrm{1}\right) \\ $$$$\frac{{dI}}{{da}}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{−{x}\mathrm{sin}\left({ax}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left({ax}\right)}{{x}}\:{dx}\:+\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left({ax}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:{dx}\:; \\ $$$$\frac{{dI}}{{da}}\:=\:−\frac{\pi}{\mathrm{2}}\:+\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left({ax}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:{dx}; \\ $$$$\frac{{dI}}{{da}}\left({a}\rightarrow+\mathrm{0}\right)\:=\:−\frac{\pi}{\mathrm{2}};\:\left(\mathrm{2}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {I}}{{da}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left({ax}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:{I}; \\ $$$$\frac{{d}^{\mathrm{2}} {I}}{{da}^{\mathrm{2}} }\:−\:{I}\:=\:\mathrm{0}\:\Rightarrow\:{I}\left({a}\right)\:=\:{C}_{\mathrm{1}} {e}^{{a}} \:+\:{C}_{\mathrm{2}} {e}^{−{a}} ; \\ $$$$\left(\mathrm{1}\right),\left(\mathrm{2}\right)\:\Rightarrow\:\begin{cases}{{C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \:=\:\frac{\pi}{\mathrm{2}}}\\{{C}_{\mathrm{1}} \:−\:{C}_{\mathrm{2}} \:=\:−\frac{\pi}{\mathrm{2}}}\end{cases}\:\Rightarrow\:\begin{cases}{{C}_{\mathrm{1}} \:=\:\mathrm{0}}\\{{C}_{\mathrm{2}\:} =\:\frac{\pi}{\mathrm{2}}}\end{cases}\:; \\ $$$${I}\left({a}\right)\:=\:\frac{\pi}{\mathrm{2}}\:{e}^{−{a}} ; \\ $$$${I}\:=\:{I}\left(\mathrm{1}\right)\:=\:\frac{\pi}{\mathrm{2}{e}}\: \\ $$

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