Does the generalised integral $$\int_{0}^{\infty}\frac{e^{\arctan(x)}-1}{x \sqrt x}\,dx$$

converge or diverge?

The first thing I do is divide it into two integrals

$\int_{0}^{A}\frac{e^{\arctan(x)}-1}{x \sqrt x}\,dx$ + $\int_{A}^{\infty}\frac{e^{\arctan(x)}-1}{x \sqrt x}\,dx$.

And then I would want to utilize something like $\frac{e^{x}-1}{x}+\frac{1}{\sqrt x}$, where $\frac{1}{\sqrt x}$ is my $g(x)$.

I think I want to use $\int_{0}^{A}\frac{dx}{\sqrt x}$ and then calculate $\frac{f(x)}{g(x)}$ and then $\int_{A}^{\infty}\frac{e^{\arctan(x)}-1}{x \sqrt x}\,dx$.

But at this point I am a bit stuck with the calculations.


Splitting the integral in two ranges, each including only one of the problematic points is wise. We can easily show that both of these integrals are finite as follows:

Consider the tail of the integral extending to infinity. We know that $\arctan(x)<\pi/2$ and thus we immediately conclude that

$$\int_{A}^{\infty}\frac{e^{\arctan x}-1}{x\sqrt{x}}dx<\int_{A}^{\infty}\frac{e^{\pi/2}-1}{x\sqrt{x}}dx=2\frac{e^{\pi/2}-1}{A^{1/2}}<\infty$$

For the first part finding a suitable bound is a bit less intuitive, but it's not that hard to tell that we need to put an upper bound on the function $f(x)=(e^x-1)/x$ which is constant in the vicinity of the origin. We take the derivative of $f$


We examine the sign of $g$. The derivative of this function is $g'(x)=xe^x$ for $x>0$ and therefore we conclude that $g$ is increasing and thus

$$g(x)> g(0)\Rightarrow f'(x)> 0 ~~\forall~~x>0 $$

We finally have concluded that $f$ is itself and increasing function, and thus we have established that an upper bound for it is given by it's value at the rightmost boundary. Since $f(x)<f(A)$ and $\arctan x<x$ we get the estimate

$$\int_{0}^{A}\frac{e^{\arctan x}-1}{x\sqrt{x}}dx<\int_{0}^{A}\frac{e^x-1}{x\sqrt{x}}dx<\frac{e^A-1}{A}\int_{0}^A\frac{dx}{\sqrt{x}}=2\frac{e^A-1}{\sqrt{A}}<\infty$$

and the integral converges to a finite value.

  • 1
    $\begingroup$ Thanks a lot, I understand everything now, had some problems with g I realised $\endgroup$ – PythonDaniel Sep 15 '20 at 22:47

What you could also have done is to check the integrand at the bounds.

Cloase to $x=0$ $$\frac{e^{\arctan(x)}-1}{x \sqrt x}=\frac 1{\sqrt x}\left(1+\frac{x}{2}+O\left(x^2\right) \right)$$ so, no problem at the lower bound.

For large values of $x$ $$\frac{e^{\arctan(x)}-1}{x \sqrt x}=\frac 1{x\sqrt x}\left(\left(e^{\pi /2}-1\right)-\frac{e^{\pi /2}}{x}+O\left(\frac{1}{x^2}\right) \right)$$ and still no problem.


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