Should $p.v.\int_0^1 x\cot(\pi x)dx$ converge?

Question

If I consider the Cauchy principal value integral of the function $x\cot(\pi x)$ can I expect convergence? For example, ordinarily the following integrals,

$$\int_0^1x\cot(\pi x)dx,$$

or

$$\int_0^{1.5}x\cot(\pi x)dx,$$

do not converge. However,

$$p.v.\int_0^{1.5}x\cot(\pi x)dx=\lim_{\varepsilon\to0^+}\left(\int_0^{1-\varepsilon}x\cot(\pi x)dx+\int_{1+\varepsilon}^{1.5}x\cot(\pi x)dx\right)\approx 0.33,$$ which I evaluated using Mathematica. I've done a bit of experimenting with Mathematica about the point $x=1$, and I get that

Integrate[x Cot[Pi x], {x, 0, 1}, PrincipalValue -> True]

evaluates to $-\infty$.

Can I not make the p.v. work like follows, $$p.v.\int_0^1x\cot(\pi x)dx=\lim_{\varepsilon\to0^+}\left(\int_0^{1-\varepsilon}x\cot(\pi x)dx+\int_{1}^{1+\varepsilon}x\cot(\pi x)dx\right),$$

or is there another way to achieve what I want? Or maybe Mathematica is right, and this definite integral does not exist?

Somehow I think it should converge because $\varepsilon>0$. So any negative massive contribution from the left would be cancelled by a positive contribution of the same (same ?) magnitude from the right? Hence, it would not diverge, but converge.

Update 1

I decided to expand $x\cot(\pi x)$ about the point $x=1$, which gave the Laurent series

$$L(x)=\frac{1}{\pi (x-1)}+\frac{1}{\pi }-\frac{1}{3} \pi (x-1)+O((x-1)^2).$$

The function $L$ seems to behave about the point $x=1$ given the context I present. For example,

$$\lim_{\varepsilon\to 0^+}L(1-\varepsilon)+L(1+\varepsilon)=\lim_{\varepsilon\to0^+}\frac{1}{-\pi\varepsilon}+\frac{1}{\pi\varepsilon}+\frac{2}{\pi}=\frac{2}{\pi}.$$

Update 2

I think the problem with update 2 in relation to my question is that I want to perform $\int_0^1$, not $\int_{-a}^a$. So I'm along the lines of thinking the $p.v.$ of my integrand does not converge over $[0,1]$...

Wait a minute, why wouldn't your first two integrals converge? They are definite and the functions are continuous. In fact, they can be evaluated through integration by parts — imranfat, Nov 13 '16 at 00:11
For starters, it would be helpful to compare the limit as x goes to 0 and 1 of $x \cot { \pi x}$. Since one is finite and the other is not, that tells you everything you need to know here. — Paul, Nov 13 '16 at 00:34
cough you meant the Laurent series at that point. And, just a thought, but contour integrals could be of use. — Simply Beautiful Art, Nov 13 '16 at 00:34
Yes, Laurent series. Residue calculus? Hmm, yes I could try that. I thought the p.v. might be able to handle it. — pshmath0, Nov 13 '16 at 00:34
Well, I mean, the principal value of $T$ is usually given by the average of the left and right limits, or the constant term in the Laurent series. — Simply Beautiful Art, Nov 13 '16 at 00:35
You could consider the similar problem: $$P.V.\int_0^1\frac1xdx$$since it behaves similarly. — Simply Beautiful Art, Nov 13 '16 at 00:40
Thanks, but I'm not sure how that helps. The expansion is $1/x$ and the p.v. about 0 is 0. — pshmath0, Nov 13 '16 at 00:44
Note the only problem with Laurent series is the $\frac1{\pi(x-1)}$, in other words, we want to take $$P.V.\int_0^1\frac1{x-1}dx$$ — Simply Beautiful Art, Nov 13 '16 at 01:05

Should $p.v.\int_0^1 x\cot(\pi x)dx$ converge?

0 Answers0