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Bessel Function of Zeroth Order is defined as \[J_{0}(x) = \frac{2}{\pi}\cdot \int_{0}^{\frac{\pi}{2}} \cos (x\cdot \sin \theta)\cdot d\theta \]

(a) Prove that \[ \int_{0}^{\infty} e^{-ax}\cdot J_{0}(x)\cdot dx = \dfrac{1}{\sqrt{1+a^{2}}}\]

(b) Prove that \[ \int_{0}^{\infty} \dfrac{\sin ax}{x}\cdot J_{0}\cdot dx = \left\{ \begin{array}{rl} \frac{\pi}{2} & \forall\ a \geq 1 \\ \arcsin (a) & \forall \ |a| \leq 1 \\ \frac{- \pi}{2} & \forall\ a \leq -1 \end{array} \right. \] 2

Laplace's Function is defined as follows :

\[ \phi (x) = \dfrac{2}{\sqrt{\pi}}\cdot \int_{0}^{x} e^{-t^{2}} \cdot dt\]

Prove that :

\[ \int_{0}^{\infty} [1- \phi (x)] dx = \frac{1}{\sqrt{\pi}}\]

P.S I need help solving the first question .

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TopNewestThe first problem - \[\begin{align} I &=\int_{0}^{\infty} e^{-ax} J_0(x) \mathrm{d}x\\ &=\int_{0}^{\infty} e^{-ax}\int_{0}^{\pi/2} \cos(x \sin \theta) \mathrm{d}\theta \mathrm{d}x\\ &=\dfrac{2}{\pi} \int_{0}^{\infty} \int_{0}^{\pi/2} e^{-ax} cos(x \sin \theta) \mathrm{d}\theta \mathrm{d}x\\&=\dfrac{2}{\pi} \int_{0}^{\pi/2}\int_{0}^{\infty}e^{-ax} \cos(x \sin \theta)\mathrm{d}x \mathrm{d}\theta\\ &=\dfrac{2}{\pi} \int_{0}^{\pi/2} \mathcal{L} [\cos(x \sin \theta)](a) \mathrm{d}\theta\\ &=\dfrac{2}{\pi} \int_{0}^{\pi/2} \dfrac{a}{a^2+\sin^2 \theta} \mathrm{d}\theta\\&= \dfrac{2a}{\pi} \int_{0}^{\pi/2} \dfrac{\sec^2{\theta}}{a^2\sec^2{\theta}+\tan^2{\theta}}\mathrm{d}\theta\\&\stackrel{\tan \theta \rightarrow t}{=} \dfrac{2a}{\pi}\int_{0}^{\infty} \dfrac{2}{(1+a^2)t^2+a^2}\mathrm{d}t\\ &=\dfrac{2a}{\pi} \dfrac{\pi}{2a\sqrt{1+a^2}}\\ &= \dfrac{1}{\sqrt{1+a^2}} \end{align}\]

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Oh, it really is easy, thanks .

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1. \[\begin{eqnarray} I = \int_0^\infty e^{-ax} J_0(x)\ dx & = & \frac{2}{\pi} \int_0^\infty e^{-ax}\left(\int_0^{\frac{\pi}{2}} \cos(x\sin \theta)\ d\theta\right)\ dx \\ & = & \frac{2}{\pi} \int_0^\infty \int_0^{\frac{\pi}{2}} e^{-ax} \cos(x\sin \theta)\ d\theta\ dx \\ & = & \Re\left(\frac{2}{\pi} \int_0^\infty \int_0^{\frac{\pi}{2}} e^{-ax} e^{ix\sin \theta}\ d\theta\ dx\right) \\ & = & \Re\left(\frac{2}{\pi} \int_0^\infty \int_0^{\frac{\pi}{2}} e^{-(a-i\sin \theta)x}\ d\theta\ dx\right) \\ \end{eqnarray}\] Now, by reversal of order of integration, we can proceed as \[\begin{eqnarray} I & = & \Re\left(\frac{2}{\pi} \int_0^\infty \int_0^{\frac{\pi}{2}} e^{-(a-i\sin \theta)x}\ d\theta\ dx\right) \\ & = & \Re\left(\frac{2}{\pi} \int_0^{\frac{\pi}{2}} \int_0^\infty e^{-(a-i\sin \theta)x}\ dx\ d\theta \right) \\ & = & \Re\left(\frac{2}{\pi} \int_0^{\frac{\pi}{2}} \frac{1}{a-i\sin \theta} d\theta \right) \\ & = & \Re\left(\frac{2}{\pi} \int_0^{\frac{\pi}{2}} \frac{a+i\sin \theta}{a^2+\sin^2 \theta} d\theta \right) \\ & = & \frac{2a}{\pi} \int_0^{\frac{\pi}{2}} \frac{1}{a^2+\sin^2 \theta} d\theta \\ \end{eqnarray}\]

Now, that's a simple integral which can be easily evaluated by using the following two trig identities : \[2\sin^2 \theta = 1- \cos 2\theta\] \[\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \]

\[\begin{eqnarray} I & = & \frac{2a}{\pi} \int_0^{\frac{\pi}{2}} \frac{1}{a^2+\sin^2 \theta} d\theta \\ & = & \frac{2a}{\pi(a^2+1)} \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta}{\frac{a^2}{a^2+1}+\tan^2 \theta} d\theta \\ \end{eqnarray}\]

Substituting \(t = \tan \theta \Rightarrow dt = \sec^2 \theta\ d\theta\) gives

\[\begin{eqnarray} I & = & \frac{2a}{\pi (a^2+1)} \int_0^{\infty} \frac{dt}{\frac{a^2}{a^2+1}+t^2} \\ & = & \frac{2a}{\pi (a^2+1)} \times \frac{\sqrt{a^2+1}}{a} \tan^{-1}\left(t\frac{\sqrt{a^2+1}}{a}\right) \huge{|}_0^\infty \\ \end{eqnarray}\]

And thus, \[\int_0^\infty e^{-ax} J_0(x)\ dx = \frac{1}{\sqrt{a^2+1}}\]

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2. At first look, \(\frac{\sin ax}{x}\) immediately rings a bell, and Dirchlet's Integral strikes my mind.

So, let us consider the function

\[I(\omega) = \int_0^\infty e^{-\omega x}\frac{\sin ax}{x}J_0(x)\ dx\]

And actually, we need to find the value of \(I(0)\). Taking partial derivative of the above function w.r.t \(\omega\) gives

\[I^\prime (\omega) = \int_0^\infty -x e^{-\omega x} \frac{\sin ax}{x} J_0(x)\ dx\] \[\Rightarrow I^\prime (\omega) = - \frac{2}{\pi} \int_0^\infty \int_0^{\frac{\pi}{2}} e^{-\omega x}\ \sin ax\ \cos(x\sin \theta)\ d\theta\ dx\]

The rest can easily be evaluated by proceeding with the method which I have followed in problem \(1.\)

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Yes, even I used this method !!! +1

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Ans to 1)

We have our double integral as :

\(I=\dfrac{2}{\pi} \displaystyle \int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \pi /2 }{ { e }^{ -ax }cos(xsin\theta )d\theta dx } } \)

Now since the limits are independent of \(\theta\) or \(x\) hence we can change the order of integration directly.

\(I=\dfrac{2}{\pi} \displaystyle \int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \pi /2 }{ { e }^{ -ax }cos(xsin\theta )dxd\theta } }\)

Now I am proving a general result:

\(\displaystyle \int _{ 0 }^{ \infty }{ { e }^{ -ax }cos(bx)dx } = \frac{a}{a^{2}+b^{2}}\) where \( a,b \epsilon R\)

Proof :

\(J=\displaystyle \int _{ 0 }^{ \infty }{ { e }^{ -ax }cos(bx)dx } =Re(\int _{ 0 }^{ \infty }{ { e }^{ -ax+ibx }dx } )\)

\(J=Re( \displaystyle\frac { { e }^{ -ax+ibx } }{ -a+ib }) \overset { \infty }{ \underset { 0 }{ | } } \)

\(J=\displaystyle \frac { (-{ e }^{ -ax }) }{ { a }^{ 2 }+{ b }^{ 2 } } (acos(bx)-bsin(bx))\overset { \infty }{ \underset { 0 }{ | } } \)

\(\Rightarrow J=\frac { a }{ { a }^{ 2 }+{ b }^{ 2 } }\)

Okay using this result in our question we have :

\(I=\displaystyle \dfrac { 2 }{ \pi } \int _{ 0 }^{ \infty }{ \frac { a }{ { a }^{ 2 }+\sin ^{ 2 }{ \theta } } d\theta } \)

\(I=\displaystyle \dfrac { 2a }{ \pi } \int _{ 0 }^{ \infty }{ \frac { \sec ^{ 2 }{ \theta } }{ { a }^{ 2 }\sec ^{ 2 }{ \theta } +\tan ^{ 2 }{ \theta } } d\theta }\)

Put \(\tan \theta =y \) to get :

\(I = \displaystyle \dfrac { 2a }{ \pi } \int _{ 0 }^{ \infty }{ \frac { dy }{ ({ a }^{ 2 }+1){ y }^{ 2 }+{ a }^{ 2 } } } \)

Now this integral is trivial :

\(I=\displaystyle \dfrac { 2a }{ \pi ({ a }^{ 2 }+1) } \int _{ 0 }^{ \infty }{ \frac { dy }{ { y }^{ 2 }+\frac { { a }^{ 2 } }{ { a }^{ 2 }+1 } } } \)

\(I=(\dfrac { 2a }{ \pi ({ a }^{ 2 }+1) } \frac { \sqrt { { a }^{ 2 }+1 } }{ a } \frac { \pi }{ 2 } )\)

\(I=\dfrac { 1 }{ \sqrt { { a }^{ 2 }+1 } } \)

Hence Proved.

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What we need to evaluate in the last problem is \(\displaystyle\int_{0}^{\infty} \mathrm{erfc} (x) \mathrm{d}x\). Let us find \(f(y)=\displaystyle\int_{y}^{\infty} \mathrm{erfc} (x) \mathrm{d}x\) and then evaluate \(f(0)\).

\[\begin{align}f(y)&=\int_{y}^{\infty} \mathrm{erfc} (x) \mathrm{d}x\\ &=\dfrac{2}{\sqrt{\pi}} \int_{y}^{\infty} \int_{x}^{\infty}e^{-z^2} \mathrm{d}z \ \mathrm{d}x\\&=\dfrac{2}{\sqrt{\pi}}\iint_{y<x<z} e^{-z^2} \mathrm{d}z \ \mathrm{d}x\\ &=\dfrac{2}{\sqrt{\pi}} \int_{y}^{\infty} \int_{y}^{z} \mathrm{d}x \ e^{-z^2} \mathrm{d}z\\ &=\dfrac{2}{\sqrt{\pi}} \int_{y}^{\infty} (z-y)e^{-z^2} \mathrm{d}z\\ &=\boxed{\dfrac{1}{\sqrt{\pi}}e^{-y^2}-y \cdot \mathrm{erfc} (y)}\end{align}\\ \]After opening the bracket in the penultimate step, the first term can be integrated by the substiitution \(y^2 \rightarrow t\). The second term is evidently \(y \cdot \mathrm{erfc} (y)\). Put \(y=0\) to get the answer.

NOTE :\(\mathrm{erfc} (x)\) is the complementary error function.Log in to reply

I didn't know of Error function , just looked it up . Nice method .+1

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Thank you!

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@Kishlaya Jaiswal , @megh choksi ,@Pratik Shastri ,@Ronak Agarwal ,@Pranjal Jain ,@Vraj Mehta ,@brian charlesworth ,@Kartik Sharma ,@Ameya Salankar and many more !!!

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Another way of doing Problem 2 could be evaluating \(\displaystyle\int_{0}^{\infty} \dfrac{\sin(ax) \cos(x \sin \theta)}{x}\mathrm{d}x\) using the property of the laplace transform that if \(\mathcal{L}\{f(t)\}=F(s)\) then \(\mathcal{L} \left\{\dfrac{f(t)}{t}\right\}=\displaystyle\int_{s}^{\infty} F(p) \mathrm{d}p\).

In our case, \(f(t)=\sin(at) \cos(t \sin \theta)\) and \(s \rightarrow 0\). After doing this, proceeding as in problem 1 should do.

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I had read about this Transform, but wasn't sure how to apply it, thanks . +1

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You're welcome :)

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@Pratik Shastri @Azhaghu Roopesh M could you please suggest some good source for studying Laplace Transformation!

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The wikipedia page is quite good. Plus it also has some examples.

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The Wolfram Mathworld that I looked up yesterday , and the video from Khan Academy which I watched a few days ago . This site too has some good examples .

Just try them out, it might come in handy for JEE Advanced .

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Actually, I am not at all familiar with Laplace's Transform, that's why I proceeded with considering another function and then taking partial derivative under the integral sign.

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That's okay :) Using Leibnitz's rule is what I had in mind too.

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How did you and Pratik came to know about all these ?

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@Pranav Arora @Haroun Meghaichi @Anastasiya Romanova , and all other Calculus pros' feed and definitely you'll learn all these awesome methods.

Just, hover aroundLog in to reply

@Haroun Meghaichi 's questions , they are a class apart and @Pranav Arora 's Integration skills are superb , but don't forget our own @Kishlaya Jaiswal and @Pratik Shastri , these guys are top-class too :)

Yeah, you should solveLog in to reply

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Well, Azhaghu Roopesh M , I was looking in a section of cylindrical wave theory i.e. a capacitor at high frequencies. There, I read about this Bessel function coming up. But it was a series like -

\(\displaystyle {E}_{\text{total}} = {E}_{0}{e}^{\iota \omega t}(\sum_{k=0}^{\infty}{\frac{{(x\iota/2)}^{2k}}{{k!}^{2}}})\) where \(y = \frac{r\omega}{2c}\)

And the series was told to be \({J}_{0}(x)\). Can you explain how this sum equal the integral? *Don't convert the integral to series, do the reverse.

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