Integrals in Disguise

Calculus Level 5

∫ 0 ∞ tan ⁡ − 1 x 2 1 + x 2 d x \int_{0}^{\infty} \frac{\tan ^{-1} x^{2}}{1+x^{2}} \, d x

The above integral is equal to which integral/s in the options?

∫ 0 ∞ tan ⁡ − 1 x 3 1 + x 2 d x \int_0^\infty \frac{\tan^{-1}x^3}{1+x^2}\,dx ∫ 0 ∞ cot ⁡ − 1 x 2 1 + x 2 d x \int_0^\infty \frac{\cot^{-1}x^2}{1+x^2}\,dx ∫ 0 ∞ cot ⁡ − 1 x 3 1 + x 2 d x \int_0^\infty \frac{\cot^{-1}x^3}{1+x^2}\,dx All of three integrals above None of these choices

Digvijay Singh by Digvijay Singh , 21, India

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1 solution

Mark Hennings
Jul 2, 2019

All four functions f 1 ( x ) = tan ⁡ − 1 x 2 f 2 ( x ) = tan ⁡ − 1 x 3 f 3 ( x ) = cot ⁡ − 1 x 2 f 4 ( x ) = cot ⁡ − 1 x 3 f_1(x) \; = \; \tan^{-1}x^2 \hspace{1cm} f_2(x) \; = \; \tan^{-1}x^3 \hspace{1cm} f_3(x) \; = \; \cot^{-1}x^2 \hspace{1cm} f_4(x) \; = \; \cot^{-1}x^3 have the property that f j ( x ) + f j ( x − 1 ) = 1 2 π x > 0 , j = 1 , 2 , 3 , 4 f_j(x) + f_j(x^{-1}) = \tfrac12\pi \hspace{2cm} x > 0\;,\; j = 1,2,3,4 Thus the substitution y = x − 1 y = x^{-1} gives ∫ 0 ∞ f j ( x ) x 2 + 1 d x = ∫ 0 ∞ f j ( y − 1 ) y 2 + 1 d y = ∫ 0 ∞ 1 2 π − f j ( x ) x 2 + 1 d x = 1 2 × 1 2 π ∫ 0 ∞ d x x 2 + 1 = 1 8 π 2 \begin{aligned} \int_0^\infty \frac{f_j(x)}{x^2+1}\,dx & = \; \int_0^\infty \frac{f_j(y^{-1})}{y^2+1}\,dy \; = \; \int_0^\infty \frac{\frac12\pi - f_j(x)}{x^2+1}\,dx \\ & = \; \tfrac12\times \tfrac12\pi \int_0^\infty \frac{dx}{x^2+1} \; = \; \tfrac18\pi^2 \end{aligned} for j = 1 , 2 , 3 , 4 j=1,2,3,4 , so all four integrals are the same.

3 Helpful 0 Interesting 7 Brilliant 0 Confused

Mark Henning's solution is the better solution. Here is a brute force solution.

∫ 0 ∞ tan ⁡ − 1 ( x 2 ) x 2 + 1 d x = ∫ 0 ∞ cot ⁡ − 1 ( x 2 ) x 2 + 1 d x = π 2 8 \int_0^{\infty } \frac{\tan ^{-1}\left(x^2\right)}{x^2+1} \, dx = \int_0^{\infty } \frac{\cot ^{-1}\left(x^2\right)}{x^2+1} \, dx = \frac{\pi ^2}{8}

∫ 0 ∞ tan ⁡ − 1 ( x 3 ) x 2 + 1 d x = G 5 , 5 4 , 5 ( 1 ∣ 1 6 , 1 2 , 1 2 , 5 6 , 1 1 6 , 1 2 , 1 2 , 5 6 , 0 ) 16 π 2 = 2 π 4 16 π 2 = π 2 8 \int_0^{\infty } \frac{\tan ^{-1}\left(x^3\right)}{x^2+1} \, dx = \frac{G_{5,5}^{4,5}\left(1\left| \begin{array}{c} \frac{1}{6},\frac{1}{2},\frac{1}{2},\frac{5}{6},1 \\ \frac{1}{6},\frac{1}{2},\frac{1}{2},\frac{5}{6},0 \\ \end{array} \right.\right)}{16 \pi ^2} = \frac{2 \pi ^4}{16 \pi ^2} = \frac{\pi ^2}{8}

∫ 0 ∞ tan ⁡ − 1 ( x 3 ) x 2 + 1 d x = G 5 , 5 5 , 4 ( 1 ∣ 1 6 , 1 2 , 1 2 , 5 6 , 1 0 , 1 6 , 1 2 , 1 2 , 5 6 ) 16 π 2 = 2 π 4 16 π 2 = π 2 8 \int_0^{\infty } \frac{\tan ^{-1}\left(x^3\right)}{x^2+1} \, dx = \frac{G_{5,5}^{5,4}\left(1\left| \begin{array}{c} \frac{1}{6},\frac{1}{2},\frac{1}{2},\frac{5}{6},1 \\ 0,\frac{1}{6},\frac{1}{2},\frac{1}{2},\frac{5}{6} \\ \end{array} \right.\right)}{16 \pi ^2} = \frac{2 \pi ^4}{16 \pi ^2} = \frac{\pi ^2}{8}

In these cases, the G function is the Meijer G function .

A Former Brilliant Member - 1 year, 11 months ago

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