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Integration Contest - Season 5

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Post Wed Jun 18, 2014 8:49 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

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Welcome! I announce the Beginning of another Integartion Contest! I am not going to post rules again and again as most of our members are familiar with the idea of contests. For newcomers I would suggest to go through the previous contests before taking part.

Remember numbering your problems while posting. Have fun and enjoy! :D

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Post Wed Jun 18, 2014 8:59 am
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

Problem 1

\(\displaystyle \int_0^\infty \frac{e^{\cos(ax)}\cos\left(\sin (ax)+bx\right)}{c^2+x^2}dx =\frac{\pi}{2c}e^{-bc+e^{-ac}}\)

where \(\displaystyle a,b,c>0\).

Post Wed Jun 18, 2014 5:57 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
$$ \begin{align} \int_0^\infty \frac{e^{\cos(ax)}\cos\big(\sin (ax)+bx\big)}{c^2+x^2}dx &= \frac{1}{2} \ \text{Re} \int_{-\infty}^\infty \frac{e^{e^{iax}} e^{ibx}}{c^2+x^2}dx \\ &= \text{Re} \ i \pi \ \text{Res} \left[\frac{e^{e^{iaz}}e^{ibz}}{c^{2}+z^{2}}, ic \right] \\ &= \pi \frac{e^{e^{-ac}}e^{-bc}}{2c} \\ &= \frac{\pi}{2c} e^{-bc+e^{-ac}} \end{align}$$

To see that $ \displaystyle \int \frac{e^{e^{iaz}} e^{ibz}}{c^2+z^2}dz$ vanishes along the upper half of $|z|=R$ as $R \to \infty$, expand $e^{e^{iaz}}$ in a Maclaurin series, switch the order of integration and summation, and then apply Jordan's lemma. Or something like that.

Post Thu Jun 19, 2014 3:42 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Great! :) Here's another:

Problem 2

\(\displaystyle \int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx=\frac{\pi}{b}e^{-b\sqrt{a^2+c^2}}\sinh (ab)\)

Post Sat Jun 21, 2014 1:44 pm
galactus User avatar
Global Moderator
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When we see an arctan, sometimes parts is a good start.

Let $\displaystyle u=\tan^{-1}\left(\frac{2ax}{x^{2}+c^{2}}\right), \;\ dv=\sinh(bx)dx, \;\ du=\frac{-2a(x+c)(x-c)}{x^{4}+2(2a^{2}+c^{2})x^{2}+c^{4}}dx, \;\ v=\frac{-1}{b}\cos(bx)$

The uv part goes to 0 and we have remaining due to it being an even function:

$$\frac{2a}{b}\int_{-\infty}^{\infty}\frac{(x-c)(x+c)\cos(bx)}{x^{4}+2(2a^{2}+c^{2})x^{2}+c^{4}}dx$$

Use a semicircle in the UHP and consider:

$$f=\frac{2a}{b}\int_{C}\frac{(z+c)(z-c)e^{ibz}}{z^{4}+2(2a^{2}+c^{2})z^{2}+c^{4}}$$

The portion around the arc tends to 0 as $R\to \infty$.

It has poles at:

$$z=(\sqrt{a^{2}+c^{2}}+a)i, \;\ z=(\sqrt{a^{2}+c^{2}}-a)i$$

$$2\pi i \cdot Res(f, \;\ (\sqrt{a^{2}+c^{2}}+a)i)=\frac{\pi }{b}e^{-ab}e^{-b\sqrt{a^{2}+c^{2}}}$$

$$2\pi i \cdot Res(f, \;\ (\sqrt{a^{2}+c^{2}}-a)i)=\frac{\pi}{b}e^{ab}e^{-b\sqrt{a^{2}+c^{2}}}$$

sum the residues in the UHP and note the exponential identity for sinh(ab). Thus, obtaining:

$$=\frac{\pi}{b}e^{-b\sqrt{a^{2}+c^{2}}}\sinh(ab)$$

Post Sat Jun 21, 2014 3:13 pm
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Very Nice, C! This one was a little tedious. :)

Problem 3

\(\displaystyle \int_0^{\pi\over 2}\frac{\sin\left( 2a\cos^2 x\right)\cosh(a\sin 2x)}{b^2\cos^2 x+c^2 \sin^2 x}dx=\frac{\pi}{2bc}\sin\left(\frac{2ac}{b+c} \right)\)

where $b,c>0$.

Post Mon Jun 23, 2014 5:04 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Let $I$ denote the integral. Then: $$\begin{align*}
I &= \int_0^{\pi\over 2}\frac{\sin\left( 2a\cos^2 x\right)\cosh(a\sin 2x)}{b^2\cos^2 x+c^2 \sin^2 x}dx \\
&= \frac{1}{2}\int_{-\pi\over 2}^{\pi\over 2}\frac{\sin\left( 2a\cos^2 x\right)\cosh(a\sin 2x)}{b^2\cos^2 x+c^2 \sin^2 x}dx \\
&= \frac{1}{4}\int_{-\pi}^\pi \frac{\sin\left( 2a\cos^2 \frac{x}{2}\right)\cosh(a\sin x)}{b^2\cos^2 \frac{x}{2}+c^2 \sin^2 \frac{x}{2}}dx \\
&= \frac{1}{4}\int_{-\pi}^\pi \frac{\sin\left( a(1+\cos x)\right)\cosh(a\sin x)}{b^2\left(\frac{1+\cos x}{2} \right)+c^2 \left(\frac{1-\cos x}{2} \right)}dx \\
&= \frac{1}{2} \text{Re}\int_{-\pi}^\pi \frac{\sin\left( a+a e^{ix}\right)}{b^2+c^2+(b^2-c^2)\cos(x)}dx
\end{align*}$$

Now substitute $z=e^{ix}$: $$\begin{align*}
I &= \frac{1}{2} \text{Re}\int\limits_{|z|=1} \frac{\sin\left( a+a z\right)}{b^2+c^2+(b^2-c^2)\frac{z+z^{-1}}{2}}\frac{dz}{iz} \\
&= \text{Re}\int\limits_{|z|=1} \frac{z\sin(a+az)}{2z(b^2+c^2)+(b^2-c^2)(z^2+1)}\frac{dz}{iz} \\
&= \frac{1}{b^2-c^2}\text{Im}\int\limits_{|z|=1} \frac{\sin(a+az)}{\left( z+\frac{b+c}{b-c}\right)\left( z+\frac{b-c}{b+c}\right)}dz
\end{align*}$$

Using Residue Theorem, $$\begin{align*}I&= \frac{1}{b^2-c^2}\text{Im}\left( 2\pi i \text{ Res}_{z=-\frac{b-c}{b+c}}\frac{\sin(a+az)}{\left( z+\frac{b+c}{b-c}\right)\left( z+\frac{b-c}{b+c}\right)} \right) \\
&= 2\pi \frac{1}{b^2-c^2} \frac{1}{\frac{4bc}{b^2-c^2}}\sin\left(\frac{2ac}{b+c} \right) \\
&= \boxed{\displaystyle \frac{\pi}{2bc}\sin\left(\frac{2ac}{b+c} \right)}
\end{align*}$$

Post Mon Jun 23, 2014 6:00 am
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

Problem 4

\(\displaystyle \begin{align*} \int_0^\infty \frac{\sin\left( 2\cos^2 x\right)\cosh(\sin 2x)}{1+x^2}dx &=\frac{\pi}{2}\sin\left(1+\frac{1}{e^2} \right)
\end{align*}\)

Post Mon Jun 23, 2014 5:07 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
Notice that the magnitude of $\sin (1+e^{2iz})= \sin(1+e^{2i(x+iy)}) = \sin \Big( 1+e^{-2y} \cos 2x + i e^{-2y} \sin 2x \Big)$ is never large in the upper half-plane since $e^{-2y} \sin 2x$ is never large in the upper half-plane.

So $\displaystyle \int \frac{\sin (1+e^{2iz})}{1+z^{2}} \ dz$ vanishes along the upper half of $|z|=R$ as $R \to \infty$.


And therefore,

$$ \begin{align} \int_{0}^{\infty} \frac{\sin (2 \cos^{2} x) \cosh(\sin 2x)}{1+x^{2}} \ dx &= \frac{1}{2} \ \text{Re} \int_{-\infty}^{\infty} \frac{\sin(1+e^{2ix})}{1+x^{2}} \ dx \\ &=\pi \ \text{Re} \ i \ \text{Res} \left[ \frac{\sin(1+e^{2iz})}{1+z^{2}}, i\right] \\ &= \frac{\pi}{2} \sin \left(1 + \frac{1}{e^{2}} \right) \end{align}$$

Post Mon Jun 23, 2014 5:25 pm

Posts: 44
Do you have some ideas about solving this problem without contour integration? :roll:

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