Evaluation of integrals

[Tsakanikas Nickos]

Evaluate the integral

• \[ \displaystyle \int_{0}^{2\pi} \frac{1}{\sin t +\cos t +2} \mathrm{d}t \]
• Let \( \gamma_{1} \) be the negatively oriented circle of centre \( i \) and radius \( \frac{1}{2} \) and let \( \gamma_{2} \) be the positively oriented circle of centre \( -i \) and radius \( \frac{1}{2} \). If \( \gamma = \gamma_{1} + \gamma_{2} \) is the sum of these two curves, then evaluate the integral
  
  
  \[ \displaystyle \int_{\gamma} \frac{ \sin z + z{e}^z }{ (z+i)^{2015} } \mathrm{d}z \]

[Papapetros Vaggelis]

Hi Nickos. I would like to share with you some thoughts on the first integral.

I tried to calculate the indefinite integral \(\displaystyle{\int \dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t}\)

by using real analysis-methods. The usual substitution is \(\displaystyle{u=\tan\,\dfrac{t}{2}}\) and then :

\(\displaystyle{\begin{aligned} \int \dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t&=\int \dfrac{1}{\displaystyle{\dfrac{2\,u}{1+u^2}+\dfrac{1-u^2}{1+u^2}+2}}\,\dfrac{2}{1+u^2}\,\mathrm{d}u\\&=\int \dfrac{1+u^2}{2\,u+1-u^2+2+2\,u^2}\,\dfrac{2}{1+u^2}\,\mathrm{d}u\\&=\int \dfrac{2}{u^2+2\,u+3}\,\mathrm{d}u\\&=\int \dfrac{2}{2+\left(u+1\right)^2}\,\mathrm{d}u\\&=\int \dfrac{1}{1+\left(\dfrac{u+1}{\sqrt{2}}\right)^2}\,\mathrm{d}u\\&=\sqrt{2}\,\arctan\,\left(\dfrac{u+1}{\sqrt{2}}\right)+c\\&=\sqrt{2}\,\arctan\,\left(\dfrac{1+\tan\,\dfrac{t}{2}}{\sqrt{2}}\right)+c\,,c\in\mathbb{R}\end{aligned}}\)

Now, by writing

\(\displaystyle{\int_{0}^{2\,\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t=\left[\sqrt{2}\,\arctan\,\left(\dfrac{1+\tan\,\dfrac{t}{2}}{\sqrt{2}}\right)\right]_{0}^{2\,\pi}=0}\)

we have a contradiction, since the function \(\displaystyle{t\mapsto \dfrac{1}{\sin\,t+\cos\,t+2}\,,t\in\left[0,2\,\pi\right]}\) is continuous, non-zero everywhere

and \(\displaystyle{\dfrac{1}{\sin\,0+\cos\,0+2}=\dfrac{1}{3}>0}\), so : \(\displaystyle{\int_{0}^{2\,\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t>0}\) .

The problem arises from the fact that the function \(\displaystyle{t\mapsto \arctan\,\left(\dfrac{1+\tan\,\dfrac{t}{2}}{\sqrt{2}}\right)}\) is not well defined

at \(\displaystyle{t=\pi}\) .

Question: Can we calculate this integral by using real-analysis methods ?

[Tolaso J Kos]

Good morning Nickos. Here is a solution without contour integration.

\begin{align*}
\int_{0}^{2\pi} \frac{{\rm d}x}{\cos x+ \sin x +2}&=\int_{-\pi}^{\pi}\frac{{\rm d}x}{\cos x + \sin x +2} \\
&\overset{u=\tan \frac{x}{2}}{=\! =\! =\! =\!}\bigintss _{-\infty}^{\infty} \frac{\frac{2}{1+t^2}}{\frac{1-t^2}{1+t^2}+ \frac{2t}{1+t^2}+2}\, {\rm d}t \\
&= \bigintss_{-\infty}^{\infty} \frac{{\rm d}t}{1+ \left ( \frac{1+t}{\sqrt{2}} \right )^2}\\
&\overset{u=\frac{1+t}{\sqrt{2}}}{=\! =\! =\!}\sqrt{2}\int_{-\infty}^{\infty}\frac{{\rm du}}{1+u^2}=\sqrt{2}\pi
\end{align*}

I also have a sol. using contour integration but I have to run it again in case I have not done anything wrong with the calculations.

[Tsakanikas Nickos]

Vaggelis, I think that the problem arises from the fact that the transformation $u=\tan \frac{t}{2}, \; t \in [0, 2\pi]$ needed to perform the change of variables is not even defined at $t=\pi$, whereas this transformation ought to be at least differentiable with integrable derivative! The solution given by Tolaso overcomes exactly this problem and therefore leads to a solution using real-analysis methods.

[Papapetros Vaggelis]

Here is a proof about the equality

\(\displaystyle{I_1=\int_{0}^{2\,\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t=\int_{-\pi}^{\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t=I_{2}}\) .

Proof

\(\displaystyle{I_1=\int_{0}^{\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t+\int_{\pi}^{2\,\pi} \dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t\,\,(I)}\)

\(\displaystyle{I_2=\int_{0}^{\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t+\int_{-\pi}^{0}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t}\) .

For the integral \(\displaystyle{\int_{-\pi}^{0}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t}\), we make the substitution \(\displaystyle{u=t+2\,\pi}\) and we get :

\(\displaystyle{\begin{aligned} \int_{-\pi}^{0}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t&=\int_{\pi}^{2\,\pi}\dfrac{1}{\sin\,(u-2\,\pi)+\cos\,(u-2\,\pi)+2}\,\mathrm{d}u\\&=\int_{\pi}^{2\,\pi}\dfrac{1}{\sin\,u+\cos\,u+2}\,\mathrm{d}u\\&=\int_{\pi}^{2\,\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t\end{aligned}}\)

Therefore :

\(\displaystyle{\begin{aligned} I_2&=\int_{0}^{\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t+\int_{-\pi}^{0}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t\\&=\int_{0}^{\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t+\int_{\pi}^{2\,\pi}\dfrac{1}{\sin\,t+\cos\,t+2}\,\mathrm{d}t\\&\stackrel{(I)}{=}I_{1}\end{aligned}}\)