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# Some Good Prove That Questions

1

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 .

Note by Azhaghu Roopesh M
1 year, 11 months ago

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The 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} · 1 year, 11 months ago

Oh, it really is easy, thanks . · 1 year, 11 months ago

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 $$1.$$ · 1 year, 11 months ago

Yes, even I used this method !!! +1 · 1 year, 11 months ago

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, $\int_0^\infty e^{-ax} J_0(x)\ dx = \frac{1}{\sqrt{a^2+1}}$ · 1 year, 11 months ago

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. · 1 year, 11 months ago

I didn't know of Error function , just looked it up . Nice method .+1 · 1 year, 11 months ago

Thank you! · 1 year, 11 months ago

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. · 1 year, 11 months ago

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. · 1 year, 8 months ago

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. · 1 year, 11 months ago

@Pratik Shastri @Azhaghu Roopesh M could you please suggest some good source for studying Laplace Transformation! · 1 year, 11 months ago

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 . · 1 year, 11 months ago

Thanks $$:)$$ · 1 year, 11 months ago

The wikipedia page is quite good. Plus it also has some examples. · 1 year, 11 months ago

Ok! thanks $$:)$$ · 1 year, 11 months ago

You're welcome :) · 1 year, 11 months ago

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. · 1 year, 11 months ago

That's okay :) Using Leibnitz's rule is what I had in mind too. · 1 year, 11 months ago

How did you and Pratik came to know about all these ? · 1 year, 11 months ago

Just, hover around @Pranav Arora @Haroun Meghaichi @Anastasiya Romanova , and all other Calculus pros' feed and definitely you'll learn all these awesome methods. · 1 year, 11 months ago

Yeah, you should solve @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 :) · 1 year, 11 months ago

Deviating from the topic ,I solved this proofathon calculus problems and have submitted some solutions on AOPS . One question was of JEE - $$\displaystyle \int_{0}^{1} \dfrac{x^t - 1}{logx} dx$$ which could be easily done. I posted a doubt - can we do this $$\displaystyle \int_{0}^{1} \dfrac{x^x - 1}{logx} dx$$ . But no response so all the participants of this note here can you please help. · 1 year, 11 months ago