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Integral Calculus

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25 Nov 2008 23:54:31 IST

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626

Evaluate the integral.

None

Evaluate the following:

$\int_0^\infty \dfrac{\tan^{-1}(\pi x) -\tan^{-1}x}{x}\ \mathrm{d}x$

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akki ~~ unlucky forever ~~

Blazing goIITian

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29 Nov 2008 11:36:03 IST

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$\int_0^\infty \dfrac{\tan^{-1}(\pi x) -\tan^{-1}x}{x}\ \mathrm{d}x$

let

integrating byparts

consider the second term of I_1,i.e,

put

simplifying ,we get

so,

similarly for

-

for

taking limits

weather or ,

so the limits becomes,

for 1st one , apply L-Hospitals rule 2 times , as both nu & den tends to infinity

for 2nd one,

so both limits reduces to zero,

FINALLY,

solving,

* * * * *

Anant Kumar

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29 Nov 2008 17:02:40 IST

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There is a minus sign between I₁ and I₂.

akki ~~ unlucky forever ~~

Blazing goIITian

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29 Nov 2008 17:13:36 IST

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yes,

actually their is only a typing mistake in a perticular line.

i have just written in place of

for rest of the sum , i have solved by taking

only a typing mistake, sorry for that......................................

Anant Kumar

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29 Nov 2008 17:18:44 IST

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In that case, shouldn't your answer turn out to be zero, since you have got I₁ = I₂ ?

akki ~~ unlucky forever ~~

Blazing goIITian

Joined: 11 May 2008

Posts: 491

29 Nov 2008 17:48:09 IST

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actually NO,

first of all, i also thought that ans would be zero,but their is in indeterminancy in it,

as, is in terms of t & t is releated to x as , & is in terms on x,

so at the time of evaluation of limits, we have to substitute t in terms of x

as is an indeterminant form,& can take finite value.just as it did in this sum.............

Anant Kumar

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29 Nov 2008 22:33:22 IST

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Yeah what you are saying, is quite correct. Basically, if you remember that

$\int_a^\infty f(t) \ \mathrm{d}t = \lim_{M\to \infty}\int_a^M f(t)\ \mathrm{d}t$

then probably it will be easier to see that $I_1\neq I_2$ .

However, a really short solution exists for the problem at hand. We generalize and use parametric differentiation. Let

$F(a)=\int_0^\infty \dfrac{\tan^{-1}(ax) -\tan^{-1}x}{x}\ \mathrm{d}x$

We differentiate with respect to $a$

$\dfrac{\mathrm{d}F}{\mathrm{d}a}=\int_0^\infty \dfrac{1}{1+(ax)^2}\ \mathrm{d}x$

Here I used the fact that since the integration is carried out over the variable $x$ while the derivative is w.r.t $a$ , so we can bring the derivative within the integrand.

The RHS is easy to obtain. So we have

$\dfrac{\mathrm{d}F}{\mathrm{d}a}= \dfrac{1}{a}\bigg|\tan^{-1}(ax)\bigg|_0^\infty = \dfrac{1}{a}\cdot \dfrac{\pi}{2}$

Integrating both side w.r.t. $a$ , we get

$F(a)=\dfrac{\pi}{2}\ln a + C$

We see that $F(1)=0$ , and so $C=0$ . Thus,

$\boxed{F(a)=\dfrac{\pi}{2}\ln a}$

Hence, the required integral

$F(\pi)=\dfrac{\pi}{2}\ln \pi$

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