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

	Answered by des_ last updated on 12/Nov/23
	$∫_0 ^∞ ((cos x)/(x^2 +1)) dx = I; I(a) = ∫_0 ^∞ ((cos(ax))/(x^2 +1)) dx, a >0; I(a→+0) = ∫_0 ^∞ (1/(x^2 +1)) dx = (π/2); (1) (dI/da) = ∫_0 ^∞ ((−xsin(ax))/(x^2 +1)) dx = −∫_0 ^∞ ((sin(ax))/x) dx + ∫_0 ^∞ ((sin(ax))/(x(x^2 +1))) dx ; (dI/da) = −(π/2) + ∫_0 ^∞ ((sin(ax))/(x(x^2 +1))) dx; (dI/da)(a→+0) = −(π/2); (2) (d^2 I/da^2 ) = ∫_0 ^∞ ((cos(ax))/(x^2 +1)) dx = I; (d^2 I/da^2 ) − I = 0 ⇒ I(a) = C_1 e^a + C_2 e^(−a) ; (1),(2) ⇒ { ((C_1 + C_2 = (π/2))),((C_1 − C_2 = −(π/2))) :} ⇒ { ((C_1 = 0)),((C_(2 ) = (π/2))) :} ; I(a) = (π/2) e^(−a) ; I = I(1) = (π/(2e))$
	$\int_{0}^{\infty} \frac{\cos x}{x^{2} + 1} d x = I;$ $I (a) = \int_{0}^{\infty} \frac{\cos (a x)}{x^{2} + 1} d x, a > 0;$ $I (a \to + 0) = \int_{0}^{\infty} \frac{1}{x^{2} + 1} d x = \frac{π}{2}; (1)$ $\frac{d I}{d a} = \int_{0}^{\infty} \frac{- x \sin (a x)}{x^{2} + 1} d x = - \int_{0}^{\infty} \frac{\sin (a x)}{x} d x + \int_{0}^{\infty} \frac{\sin (a x)}{x (x^{2} + 1)} d x;$ $\frac{d I}{d a} = - \frac{π}{2} + \int_{0}^{\infty} \frac{\sin (a x)}{x (x^{2} + 1)} d x;$ $\frac{d I}{d a} (a \to + 0) = - \frac{π}{2}; (2)$ $\frac{d^{2} I}{{d a}^{2}} = \int_{0}^{\infty} \frac{\cos (a x)}{x^{2} + 1} d x = I;$ $\frac{d^{2} I}{{d a}^{2}} - I = 0 \Rightarrow I (a) = C_{1} e^{a} + C_{2} e^{- a};$ $(1), (2) \Rightarrow {\begin{cases} C_{1} + C_{2} = \frac{π}{2} \\ C_{1} - C_{2} = - \frac{π}{2} \end{cases} \Rightarrow {\begin{cases} C_{1} = 0 \\ C_{2} = \frac{π}{2} \end{cases};$ $I (a) = \frac{π}{2} e^{- a};$ $I = I (1) = \frac{π}{2 e}$
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