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I entered this integral into Wolframalpha, and got $$\int_{0}^{\infty}\frac{1}{t}\arctan\left(\frac{t}{1+2t^2}\right)\,\mathrm dt=\frac{1}{2}\pi\log{2}.$$ But it doesn't provide step by step solution for this integral.

This integral is a bonus challenge in my Calculus class, and the professor that the key is $\arctan$. But I don't know is there any special about $$\arctan\left(\frac{t}{1+2t^2}\right),$$ so I tried some common integration method, and it doesn't work.

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    $\begingroup$ Well, do you rely entirely on WolframAlpha? What have you tried to solve it? For a hint, try to rewrite $\arctan\left(\frac{t}{1+2t^2}\right)$ in a form of $\arctan(at)\pm \arctan(bt)$. $\endgroup$ – Zacky Dec 21 '19 at 10:41
  • $\begingroup$ @Zacky Yes, I did try some integration method that I knew, but none of them works. $\endgroup$ – 3142 maple Dec 21 '19 at 10:45
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    $\begingroup$ Then can you mention some integration methods that you know and tried? Bringing context is helpful. Are you familiar with contour integration or Feynman's trick or Frullani's integral (etc)? See also: math.meta.stackexchange.com/questions/9959/… $\endgroup$ – Zacky Dec 21 '19 at 10:47
  • $\begingroup$ @Zacky Thanks, I have added some context about the question. And in the list of integration methods you listed, I only know Feynman's trick. $\endgroup$ – 3142 maple Dec 21 '19 at 10:55
  • $\begingroup$ After I googled Frullani integral, I found that the solution is quite similar to it, maybe $arctan(\frac{t}{1+2t^2})$ can be converted to $f(at)-f(bt)$. $\endgroup$ – 3142 maple Dec 21 '19 at 11:03
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Following @Zacky's hint, use $\frac{t}{1+2t^2}=\frac{2t-t}{1+2t\cdot t}$ to rewrite the integral as the Frullani integral$$\int_0^\infty\frac{\arctan(2t)-\arctan t}{t}dt=(\arctan0-\arctan\infty)\ln\frac12=\frac{\pi}{2}\ln 2.$$

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  • $\begingroup$ I found that $arctan(2t)-arctan(t)$ can be converted into another integral, than change the order of the whole double integral can make it easy to solve. Without using Frullani integral. $\endgroup$ – 3142 maple Dec 21 '19 at 11:21
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First notice that: $$\arctan\left(\frac{x}{1+2x^2}\right)=\arctan\left(\frac{2x-x}{1+2x\cdot x}\right)=\arctan(2x)-\arctan(x)$$ So the integral can be rewritten as: $$I=\int_0^\infty \frac{\arctan(2x)-\arctan x}{x}dx\overset{IBP}=\int_0^\infty \ln x\left(\frac{1}{1+x^2}-\frac{2}{1+4x^2}\right)dx$$

$$2\int_0^\infty \frac{\ln x}{1+4x^2}dx\overset{2x\to x}=\int_0^\infty \frac{\ln x-\ln 2}{1+x^2}dx$$ $$\Rightarrow I=\int_0^\infty \frac{\ln x -\ln x+\ln 2}{1+x^2}dx=\ln 2\int_0^\infty \frac{dx}{1+x^2}=\frac{\pi}{2}\ln 2$$

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    $\begingroup$ This technique can be generalised to evaluate Frullani integrals, viz.$$\begin{align}\int_{0}^{\infty}\frac{f\left(ax\right)-f\left(bx\right)}{x}dx &=\int_{0}^{\infty}\ln x\cdot\left(bf^{\prime}\left(bx\right)-af^{\prime}\left(ax\right)\right)dx\\&=\left[c\int_{0}^{\infty}\ln x\cdot f^{\prime}\left(cx\right)dx\right]_{a}^{b}\\&=\left[\int_{0}^{\infty}\ln\frac{y}{c}\cdot f^{\prime}\left(y\right)dy\right]_{a}^{b}\\&=\ln\frac{a}{b}\cdot\int_{0}^{\infty}f^{\prime}\left(y\right)dy\\&=\ln\frac{b}{a}\cdot\left(f\left(0\right)-f\left(\infty\right)\right).\end{align}$$ $\endgroup$ – J.G. Dec 21 '19 at 12:19
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By integration by parts we have

$$I=\int_0^\infty\ln x\cdot\frac{2x^2-1}{4x^4+5x^2+1}dx$$

$$=\int_0^\infty\frac{\ln x}{1+x^2}dx-\color{red}{\int_0^\infty\frac{2\ln x}{1+4x^2}dx}$$

$$\overset{\color{red}{2x\mapsto x}}{=}\int_0^\infty\frac{\ln2}{1+x^2}dx=\frac{\pi}{2}\ln2$$

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I found another way to solve this is to make the integral to be a double integral, and change the order. But the key still is $\arctan(\frac{t}{1+2t^2})=\arctan(2t)-\arctan(t)$.

$$\int_0^\infty\frac{\arctan(\frac{t}{1+2t^2})}{t}=\int_0^\infty\frac{\arctan(2t)-\arctan t}{t}dt=\int_0^\infty \frac{1}{t} \int_1^2 \frac{t}{1+(yt)^2} dydt$$ $$=\int_1^2 \int_0^\infty \frac{1}{1+(yt)^2} dtdy=\int_1^2 \frac{1}{y} \arctan(\infty)-\arctan(0) dy=\frac{\pi}{2}ln2$$

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    $\begingroup$ Since you seem interested in this, if you would like you can try to solve these too: $$\int_0^\infty \frac{\arctan\left(\frac{2x}{1+x^2}\right)}{x}dx=\pi\ln(1+\sqrt 2)$$ $$\int_0^\infty \frac{\arctan\left(\frac{2x}{1+2x^2}\right)}{x}dx=\frac{\pi}{2}\ln(2+\sqrt 3)$$ $$\int_0^\infty \frac{\arctan\left(\frac{x}{1+x^2}\right)}{x}dx=\pi \ln\varphi ,\ \varphi=\frac{1+\sqrt 5}{2}$$ For start think of a way to find the "key" in those cases. $\endgroup$ – Zacky Dec 21 '19 at 13:04

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