The value of the integral \(\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}}\) is
π
Integral Evaluation Explained
The question asks for the value of the definite integral of the function \(\frac{1}{{1 + {x^2}}}\) over the entire real line, from negative infinity to positive infinity. This is a common integral in calculus.
The integral we need to evaluate is given by: \[\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}}\]
Antiderivative of the Function
To solve this definite integral, we first need to find the antiderivative of the integrand \(\frac{1}{{1 + {x^2}}}\).
- We know that the derivative of the inverse tangent function, \(\arctan(x)\) (also written as \(\tan^{-1}(x)\)), is \(\frac{1}{{1 + {x^2}}}\).
- Therefore, the antiderivative of \(\frac{1}{{1 + {x^2}}}\) with respect to \(x\) is \(\arctan(x)\).
Evaluating the Definite Integral at Infinity
Now, we evaluate the definite integral using the fundamental theorem of calculus for improper integrals. This involves taking limits as \(x\) approaches infinity and negative infinity.
The evaluation proceeds as follows: \[\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}} = \left[ {\arctan(x)} \right]_{ - \infty }^\infty \] This can be written as: \[\lim_{b \to \infty} \arctan(b) - \lim_{a \to -\infty} \arctan(a)\]
- Limit as \(x\) approaches positive infinity: As \(x\) approaches positive infinity, the value of \(\arctan(x)\) approaches \(\frac{\pi}{2}\). \[\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}\]
- Limit as \(x\) approaches negative infinity: As \(x\) approaches negative infinity, the value of \(\arctan(x)\) approaches \(-\frac{\pi}{2}\). \[\lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}\]
Calculating the Final Value of the Integral
Substitute these limit values back into the expression for the definite integral: \[\frac{\pi}{2} - \left( -\frac{\pi}{2} \right)\] \[= \frac{\pi}{2} + \frac{\pi}{2}\] \[= \pi\]
Thus, the value of the integral \(\mathop \smallint \limits_{ - \infty }^\infty \frac{{dx}}{{1 + {x^2}}}\) is \(\pi\). This integral is a classic result in calculus, often associated with the calculation of areas related to the Cauchy distribution or probability.
Important Questions from Definite Integrals
The Legendre polynomials P n(x), n = 0, 1, 2, ..., satisfying the orthogonailty condition \(\int_{{\rm{ - 1}}}^{\rm{1}} {{{\rm{P}}_{\rm{n}}}\left( {\rm{x}} \right){{\rm{P}}_{\rm{m}}}} \left( {\rm{x}} \right){\rm{dx}}\,{\rm{ = }}\,\frac{{\rm{2}}}{{{\rm{2n + 1}}}}{{\rm{\delta }}_{{\rm{nm}}}}\) on the interval [-1, +1], may be defined by the Rodrigues formula P n(x) = \(\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}{\rm{n!}}}}\frac{{{{\rm{d}}^{\rm{n}}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} \right)^{\rm{n}}}\) . The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is
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