# Interesting real integrals from $I=\oint\limits_{C}\sin(\frac{1}{z})\,\mathrm{dz} = 2\pi i$

I have separated and equated Real and Imaginary Parts of the following equation by plugging $$\mathrm{z}=e^{it}$$ :

$$I=\oint\limits_{C}\sin\frac{1}{z}\,\mathrm{dz} = 2\pi i,$$
where $$C$$ is the boundary of unit circle centered at origin.

which gives $$K=\int_{-\pi}^{+\pi}e^{it}\sin(e^{-it})\mathrm{dt} = 2\pi$$

simplifying gives the following:-

$$\int_{0}^{\frac{\pi}{2}}\cos(t)\sin(\cos(t))\cosh(\sin(t))+\sin(t)\cos(\cos(t))\sinh(\sin(t))\mathrm{dt} = \frac{\pi}{2}$$

the first and second part of the above integral translates to $$I_1$$ and $$I_2$$ respectively under a simple substitution for each.

$$I_1 = \int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\mathrm{dx}\\ I_2 = \int_{0}^1 \cos(x)\sinh(\sqrt{1-x^2})\mathrm{dx}$$

which implies, $$I_1 + I_2 = \frac{\Large\pi}{2}$$

Interestingly, I have found that $$I_1$$ and $$I_2$$ are equal (using a calculator), but stuck proving them. Any ideas how they are equal or finding another method to calculate $$I_1$$ or $$I_2$$?

EXTRA: some integrals that show up but eventually cancel out( as they are odd functions, $$f(x)=f(-x)$$ ) in calculating Imaginary part of equation $$K$$:-

$$I_3 = \int_{0}^1 \sin(x)\cosh(\sqrt{1-x^2})\mathrm{dx}\\ I_4 = \int_{0}^1 \sinh(x)\cos(\sqrt{1-x^2})\mathrm{dx}$$

however $$I_3$$ and $$I_4$$ are not equal and their numerical values are approx. 0.584 and 0.418 , can $$I_3$$ and $$I_4$$ be calculated in closed form?

• This integral is equal to the residue of $\sin(1/z)$ times $2\pi i$. So find the Laurent series and the coefficient $a_{-1}$ and you will get your final result. Dec 6 '20 at 17:10
• I started o the editing, but I wasn't sure if it's what you had in mind. I also didn't want to impose my tastes on you. I'll mention a few things. Use display mode to offset formulas. You can get adjustable-size parentheses, brackets, and so by using \left and right. For example $\sin\left\sqrt{1-x^2}\right)$ gives $\sin\left(\sqrt{1-x^2}\right)$. Dec 6 '20 at 17:11
• @Daniel equation I already contains that, the question is to find I1,I2,I3,I4 Dec 6 '20 at 17:12
• @saulspatz thanks for the suggestions! Dec 6 '20 at 17:14

\begin{align} I_1&=\int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\,dx\\ &\stackrel{x\to\sqrt{1-t^2}}{=} \int_{0}^1 \cosh(\sqrt{1-t^2})\sin(t)\frac{t\,dt}{\sqrt{1-t^2}}\\ &=-\int_{0}^1 \sin(t)\,d(\sinh(\sqrt{1-t^2}))\\ &=\left[-\sin(t)\sinh(\sqrt{1-t^2})\right]_0^1+\int_{0}^1 \sinh(\sqrt{1-t^2})\,d (\sin(t))\\ &=\int_{0}^1 \sinh(\sqrt{1-t^2})\cos(t)\, dt=I_2. \end{align}

It follows $$I_1=I_2=\frac\pi4$$.

Applying the same method one can show $$I_3+I_4=\cosh(1)-\cos(1)$$.

• Surprisingly Mathematica cannot evaluate the integrals.
– user
Dec 8 '20 at 8:31
• yeahh, anyways nice answer! Dec 11 '20 at 11:41

The value of the integral $$I_{3} = \int_{0}^{1} \sin(x) \cosh \left( \sqrt{1-x^{2}} \right) \, \mathrm dx$$ can be expressed in terms of the cosine integral $$\operatorname{Ci}(x)$$ and the hyperbolic cosine integral $$\operatorname{Chi}(x)$$.

\begin{align} I_{3} &= \int_{0}^{\pi/2} \sin(\sin t) \cosh(\cos t) \cos(t) \, \mathrm dt \\ &= \Im \int_{0}^{\pi/2} \sinh(e^{it}) \cos (t) \, \mathrm dt \\ &= \Im \int_{C} \sinh(z) \, \frac{z+\frac{1}{z}}{2} \frac{dz}{iz} \\ &= -\frac{1}{2} \, \Re \int_{C} \left( \sinh(z) + \frac{\sinh (z)}{z^{2}} \right) \, \mathrm dz, \end{align} where $$C$$ is the portion of the unit circle in the first quadrant of the complex plane.

But since the integrand is analytic in the first quadrant, we have\begin{align} I_{3} &= - \frac{1}{2} \, \Re \int_{1}^{i} \left( \sinh(z) + \frac{\sinh (z)}{z^{2}} \right) \, \mathrm dz \\ &= - \frac{1}{2} \Re \left(\cosh (z) - \frac{\sinh (z)}{z} \Bigg|^{i}_{1} + \int_{1}^{i} \frac{\cosh (z)}{z} \, \mathrm dz \right) \\ &= - \frac{1}{2} \Re \left(\cosh (z) - \frac{\sinh (z)}{z} + \operatorname{Chi}(z) \Bigg|_{1}^{i} \right) \\ &= - \frac{1}{2} \left(\cos(1)- \sin(1) + \Re \left(\operatorname{Chi}(i) \right)- \cosh(1) + \sinh(1) - \operatorname{Chi}(1)\right) \\ &= - \frac{1}{2} \left(\cos(1)- \sin(1) + \operatorname{Ci}(1) - \cosh(1) + \sinh(1) - \operatorname{Chi}(1)\right) \\ &= \frac{1}{2} \left(\sin(1) - \cos(1) - \operatorname{Ci}(1) + \frac{1}{e} + \operatorname{Chi}(1) \right) \\ & \approx 0.58475. \end{align}

The evaluation of $$I_{4} = \int_{0}^{\pi/2} \sinh(\sin t) \cos(\cos t) \cos (t) \, \mathrm dt = \Im \int_{0}^{\pi/2} \sin(e^{it}) \cos(t) \, \mathrm dt$$ should be similar.

• Great answer !! leant something new Dec 11 '20 at 11:40