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I have the integral $$ F(s) = \int_{0}^{\infty} \frac{\arctan(sx)}{x(1+x^2)} dx$$ and am supposed to solve it by finding $F'(s)$. So we get $$ F'(s) = \int_{0}^{\infty} \frac{\partial F}{\partial s} \frac{\arctan(sx)}{x(1+x^2)} dx =...= \int_{0}^{\infty} \frac{1}{1+s^2x^2+x^2+s^2x^4} dx $$ and I don't see how I can solve this integral. Should I try another approach?

wythagoras
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Lozansky
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  • Hint: $\frac{\partial [\frac{\tan ^{-1}sx}{x(1*x^{2})}]}{\partial s}=\left ( \frac{x}{1+(sx)^{2}}\right )\left ( \frac{1}{x(1+x^{2})} \right )$ – Matematleta May 30 '15 at 14:22
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    HINT: Using residue calculus, you can show that, for $s>0$ you have $$\int_{-\infty}^{+\infty} \frac{dx}{(1+x^2)(1+x^2s^2)} = \frac{\pi}{s+1}$$ – Crostul May 30 '15 at 14:22
  • i got for the integral $\frac{1}{4} \pi \left(\log \left(1-s^2\right)+2 \tanh ^{-1}(s)\right)$ and the derivative for $s>0$ is $\frac{\pi}{2(1+s)}$ – Dr. Sonnhard Graubner May 30 '15 at 14:55
  • @Chilango Yes that is what I got, only I multiplied the parentheses and cancelled the x. Could you be more specific about your hint? – Lozansky May 30 '15 at 15:04
  • @Crostul I am not familiar with that result or residue calculus. – Lozansky May 30 '15 at 15:04
  • @Dr.SonnhardGraubner I do not see how you obtained that result. I am supposed to solve it using differentiation under integral sign. – Lozansky May 30 '15 at 15:05
  • @ Lozansky: Now cancel the x's and use partial fracton decomposition, or residues. – Matematleta May 30 '15 at 15:19

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Partial fractions: $$ \int_0^\infty \frac{1}{(1+(sx)^2)(1+x^2)}\;dx = \frac{s}{s^2-1}\int_0^\infty \frac{s}{1+(sx)^2}\;dx + \frac{1}{1-s^2}\int_0^\infty\frac{1}{1+x^2}\;dx \\ =\frac{s}{s^2-1}\;\frac{\pi}{2}+\frac{1}{1-s^2}\;\frac{\pi}{2} =\frac{1}{s+1}\;\frac{\pi}{2} $$

GEdgar
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