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Let $x > 0$. Prove that the value of the following expression doesn't depend on x

$$\int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$

Attempt:

Left: f'(x) = $\frac{1}{1+x^2}$

Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$

$= \frac{1}{(1+\frac{1}{x^2})} - \frac{1}{x^2}$

$=\frac{x^2}{1+x^2} - \frac{1}{x^2}$

$=\frac{x^4 - x^2 - 1}{1+x^2}$

Yeah I don't know what I am doing, I tried to remove the integral but failed miserably

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    You have an equation, which is not an expression and doesn't have value. Are you sure you've copied the problem correctly? – 2017-02-08
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    The = sign should be replaced by +. – 2017-02-08
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    Sorry I wrote the question wrong, yeah it should be + , thank you. – 2017-02-08

4 Answers 4

0

In the second integral, let $t=1/u$ to get

$$\int_0^{1/x}\frac1{1+t^2}\ dt=\int_x^\infty\frac1{1+(1/u)^2}\frac{du}{u^2}=\int_x^\infty\frac1{1+u^2}\ du$$

Add it to the first integral to get

$$I=\int_0^\infty\frac1{1+t^2}\ dt=\frac\pi2$$

which does not depend on $x$.

2

Note that $\int_0^x \frac{1}{1+t^2}\,dt=\arctan(x)$.

Recalling $\bbox[5px,border:2px solid #C0A000]{\arctan(x)+\arctan(1/x)=\pi/2\text{sgn(x)}}\,\,$, we see that the term of interest depends only on the sign of $x$.

From the pre-edited question

$$\arctan(x)=\arctan(1/x)\implies \arctan(x)=(\pi/4)\text{sgn}(x)\implies x=1\,\,\text{or}\,\,x=-1$$

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    I know that this equals zero? Does that prove dependency? – 2017-02-08
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    @Dr.MV Isn't it $\arctan(x)+\arctan(1/x) = \frac{\pi}{2}$? – 2017-02-08
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    @JackyChong Yes, correct. – 2017-02-08
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    @Dr.MV Could you also update your post. – 2017-02-08
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    arctan(x)+arctan(1/x)=π/2 . So yeah uhm, I was looking at tan identites, I cant find this one? Why is it pi/2 – 2017-02-08
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    @user349557 I've edited to clarify. – 2017-02-08
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    @JackyChong I've edited the typo and added to the solution. – 2017-02-08
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    What does sgn stand for? – 2017-02-08
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    @user349557 It is the [sign function](https://en.wikipedia.org/wiki/Sign_function) – 2017-02-08
1

To answer the question as-is after having been heavily edited, let for $x \gt 0\,$:

$$f(x) = \int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$

Then, using the Leibniz integral rule:

$$ f'(x) = \frac{1}{1+x^2} \,-\, \frac{1}{x^2} \cdot \frac{1}{1+ \cfrac{1}{x^2}} = \frac{1}{1+x^2} \,-\, \frac{1}{1+x^2} \,=\, 0 $$

Thus $f'(x)=0\,$, so $f(x)$ is a constant, and therefore does not depend on $x$.


P.S. Note to the OP:

Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$

This looks like you attempted to use Leibniz' rule, but misapplied it. The $-\,\frac{1}{x^2}$ derivative of the upper bound is multiplied with, not added to, the end value of the function being integrated.

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    Thank you. This was the solution I was looking for, everything else confused me. – 2017-02-08
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    @user349557 Glad it helped. For good measure, the other posted answers are all correct as well, yet this looked to me like the more direct way to get it done. – 2017-02-08
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    Wasn't saying they were not correct. Just have not learned any of the concepts in the way they did it. – 2017-02-09
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Using the $u$ sub $t' = t^{-1}$, we see $dt'= -(t')^2dt$ that \begin{align} \int^{\frac{1}{x}}_0\frac{1}{1+t^2}\ dt = -\int^x_\infty \frac{1}{1+(t')^{-2}} \frac{dt'}{(t')^2}= \int^\infty_x \frac{1}{1+(t')^2}\ dt' = \frac{\pi}{2} - \int^x_0\frac{1}{1+(t')^2}\ dt'. \end{align}

But we see that \begin{align} \frac{\pi}{2} - \int^x_0\frac{1}{1+(t')^2}\ dt'= \int^x_0\frac{1}{1+(t')^2}\ dt' \end{align} provided \begin{align} \int^x_0\frac{1}{1+(t')^2}\ dt' = \frac{\pi}{4}. \end{align} Thus, your claim is incorrect since equality holds only when $x=1$.

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    I have not learned improper integrals yet. So I have no idea how to understand in this way. – 2017-02-08
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    Jackie, $x=-1$ is also a solution. You're tacitly assuming that $x>0$ – 2017-02-08