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How does one prove that for $x>0$ $$\int_0^{1/x}{\frac{1}{t^2+1}\mathrm{d}t} + \int_0^{x} {\frac{1}{t^2+1}\mathrm{d}t}$$ is equal to a constant?

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    Your integrals are incomplete, the closing $dt$ is missing. Then perform a change of variable $t\mapsto 1/t$ in the first integral. – 2017-01-12
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    $$\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$$ is pretty trivial, geometrically. – 2017-01-12

4 Answers 4

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Use the fundamental theorem of calculus to differentiate the integrals. The second is easy; the first is \begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^{\frac{1}{x}}\frac{1}{t^2+1}\mathrm{d}t=-\frac{1}{1+x^2} \end{equation*} It is clear that this will cancel with your result from the second integral. Since the resulting derivative is zero, you must have been dealing with the constant function from the start!

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    How did you specifically differentiate the first integral? The 1/x is what's confusing me. – 2017-01-12
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    Oops, mine only works if we assume that $x>0$ (or $x<0$. I forgot the function is not defined at 0! – 2017-01-12
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    You might find this video helpful: https://www.youtube.com/watch?v=PGmVvIglZx8 – 2017-01-12
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    Aptly put. (+1) – 2017-01-12
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We know that $$\int_{0}^{\frac{1}{x}}\dfrac{1}{1+t^2}dt=\arctan\bigl(\frac{1}{x}\bigr)$$ and $$\int_{0}^{x}\dfrac{1}{1+t^2}dt=\arctan(x)$$ ( if you don't know these, use the substitution $t=\tan(u)$ and carry out the integral appropriately ) So, $$\int_{0}^{\frac{1}{x}}\dfrac{1}{1+t^2}dt+\int_{0}^{x}\dfrac{1}{1+t^2}dt=\arctan\bigl(\frac{1}{x}\bigr)+\arctan(x)=\frac{\pi}{2}$$ To explore why this trig identity is true, let $n=\arctan\bigl(\frac{1}{x}\bigr)$. Then (a rough sketch), $$\implies \tan(n)=\frac{1}{x}\implies x=\cot(n)=\tan\bigl(\frac{\pi}{2}-n\bigr)\implies \arctan(x)=\frac{\pi}{2}-n$$ $$\implies \arctan(x)+n=\frac{\pi}{2}\implies \arctan(x)+\arctan\bigl(\frac{1}{x}\bigr)=\frac{\pi}{2}$$ which is a constant.

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\begin{align} \color{red}{\int_0^{1/x}{\frac{1}{t^2+1}dt}} + \int_0^{x} {\frac{1}{t^2+1}dt}&=\color{red}{-\int_{\infty}^x\frac{dt}{1+t^2}}+\int_0^{x} {\frac{1}{t^2+1}dt}\\ &=\color{red}{\int_x^{\infty}\frac{dt}{1+t^2}}+\int_0^{x} {\frac{1}{t^2+1}dt}\\ &=\color{red}{\int_0^{\infty}\frac{dt}{1+t^2}-\int_0^x\frac{dt}{1+t^2}}+\int_0^{x} {\frac{1}{t^2+1}dt}\\ &=\int_0^{\infty}\frac{dt}{1+t^2} \end{align} where the last is free of $x$ and hence is a costant.

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Express each integral into sum of two integrals by dividing the interval of integration into two parts. Thus first integral is split as two integrals: one over $[0,1]$ and other over $[1,1/x]$. Call these as $A$ and $B$ respectively. Similarly split the second one into integrals: one over $[0,1]$ and the other over $[1,x]$. Call these $C$ and $D$ respectively. It is now easily seen that $A=C$ and using substitution $t=1/u$ we can show that $B=-D$. So the original sum of two integrals is equal to $$2A=2\int_{0}^{1}\frac{dt}{t^{2}+1}$$ which is constant (it does not involve any $x$).

A much simpler approach is the one suggested by Lutz in comments to the question but involves the use of improper Riemann integrals which the above approach avoids.