Question

If \(\lim_{a \to \infty} \frac{1}{a} \int_{a}^{\infty} \frac{x^2 + ax+1}{1+x^4} . \tan^{-1} \left(\frac{1}{x}\right) dx \) is equal to \(\frac{\pi^2}{k}\), where \(k \in \mathbb{N}\) equals to ... ?

Answer 1

\[\lim_{a \to \infty} \cfrac{1}{a} \int_{a}^{\infty} \frac{x^2 + ax+1}{1+x^4} . \tan^{-1} \left(\cfrac{1}{x}\right) dx = \cfrac{\pi ^2}{k} \]

here is a much better LaTeX for the question.

Answer 2

My bad, the limits of the integral are from 0 to \(\infty\) instead of \(\bf{a}\) to \(\infty\) .

Answer 3

This is what I've done so far, \(\tan^{-1} \left(\cfrac{1}{x}\right) = \cfrac{\pi}{4} \)

So, it becomes :

\[\lim_{0 \to \infty} \cfrac{1}{a} \int_{a}^{\infty} \frac{x^2 + ax+1}{1+x^4} . \cfrac{\pi}{4} \]

Let, \[ I = \cfrac{1}{a} \int_{0}^{\infty} \frac{x^2 + ax+1}{1+x^4} . \cfrac{\pi}{4} \]

\[ I = \cfrac{1}{a} \left[ \int_{0}^{\infty} \cfrac{x^2 + 1}{x^4 + 1} dx + a \int_{0}^{\infty} \cfrac{x dx}{1+x^4} \right] \]

I'm not sure how to go on from here.

Answer 4

I do think that there is an application of inverse trigonometric functions. I did have a look at some of the formulas : http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

But, they don't seem to fit in there. Any help will be greatly appreciated.

Answer 5

How does \(\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}\)? I think you have to evaluate the integral first.

Answer 6

One sec. I guess, I did a mistake. I'll check it out.

Answer 7

Well, we have: \(\tan^{-1} (1/x) + \cot^{-1} x = \pi /2 \)

Right?

Answer 8

But where is the \(\cot^{-1}(x)\)?

Answer 9

.

Answer 10

We do have : \(\tan^{-1} (1/x) = \cot^{-1} x \) . So, \(\tan^{-1} (1/x) + \tan^{-1} (1/x) = \pi /2 \)

So, it gives \(\tan^{-1} 1/x = \pi /4 \)

Answer 11

But \(\tan^{-1}\left(\frac{1}{2}\right)=0.463648\neq\frac{\pi}{4}=0.785398\).

Answer 12

Wait, actually I must not have generalized it. There is surely some problem with my point.

Let us put x = 1/t

We have : \(\tan^{-1} x + \cot^{-1} x = \pi /2 \)

\(\cot^{-1} t+ \cot^{-1} 1/t = \pi/2 \)

uhmm.. no! I wonder why then my book wrote the integral as :

Let \(I = \int_{0}^{\infty} \cfrac{x^2 + ax + 1 }{ 1+x^4} . \tan^{-1} 1/x dx \)
Put x = 1/t and adding we get,

\[ I = \frac{\pi}{4} \int_{0}^{\infty} \frac{ (x^2+1)+ax}{1+x^4} dx \]

Answer 13

It also stated : "using \(\tan^{-1}(1/x) + \cot^{-1} x = \pi /2 \) "

I'm also confused from where did it get to the above equation.

Answer 14

did you try lhopital ?
first you need to show that the expression is in indeterminate form

Answer 15

nvm

Answer 16

it might actually work , but we need to be careful when differentiating with respect to "a"
and changing the order of integral and derivative

Answer 17

The lower limit of integration is 0 not a?

Answer 18

\[I = \int_{0}^{\infty} \cfrac{x^2 + ax + 1 }{ 1+x^4} . \tan^{-1}( 1/x)~ dx ~~~\Large{\color{Red}{\star }}\]

Let \(x=\dfrac{1}{t}\implies dx = \dfrac{-dt}{t^2}\)

\(\Large{\color{Red}{\star }}\) becomes
\[\begin{align}I &= \int_{\infty}^0 \cfrac{1/t^2 + a/t + 1 }{ 1+1/t^4} . \tan^{-1} (t)~\frac{-dt}{t^2}\\~\\
&= \int_0^{\infty} \cfrac{t^2+at+1 }{ 1+t^4} . \tan^{-1} (t)~dt\\~\\
&=\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4} . \tan^{-1} (x)~dx~~~\Large{\color{blue}{\star}}
\end{align}\]

\(\Large{\color{Red}{\star }} + \Large{\color{blue}{\star }}\) gives
\[\begin{align}I+I&=\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4} . \left(\tan^{-1} (1/x)+\tan^{-1} (x)\right)~dx\\~\\
&=\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4} . \left(\frac{\pi}{2}\right)~dx\\~\\
\end{align}\]

Answer 19

Need to find the value of the natural number \(k\) @perl

Answer 20

this looks like one of those brilliant problems where the answer "requires" to be an integer between 0 and 999

Answer 21

is the lower limit supposed to be zero

Answer 22

Yes it was fixed later

Answer 23

thanks :)

Answer 24

Find \(k\) in below equation given that \(k\in \mathbb{N}\)

\[\lim_{a \to \infty} \cfrac{1}{a} \int_{0}^{\infty} \frac{x^2 + ax+1}{1+x^4} . \tan^{-1} \left(\cfrac{1}{x}\right) dx = \cfrac{\pi ^2}{k}\]

Answer 25

Yeah thats the complete question

Answer 26

the answer is k = 16

Answer 27

Yes

Answer 28

how in heavens name did you get that?

Answer 29

as pointed out by mathslover earlier, we can evaluate \[\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4}\,dx\]
by splitting like this
\[\int_0^{\infty} \cfrac{x^2+1 }{ 1+x^4}\,dx +a\int_0^{\infty} \cfrac{x }{ 1+x^4}\,dx \]

@alekos

Answer 30

If we're clever enough, we don't need to evaluate the first integral

Answer 31

second integral evaluates nicely by trivial substitution \(u=x^2\)

Answer 32

I wonder how did we replace t by x before that \(\color{blue}{\star}\) equation ? Shouldn't it be replaced by 1/x too? I know that would give the same equation that we started from but replacing t by x doesn't make sense to me :(

Answer 33

How do you know the first integral is 0?

Answer 34

\[\int_a^bf(t)\,dt = \int_a^bf(x)\,dx\]

@mathslover

Answer 35

x or t or something, it doenst matter
the variable in definite integral is "dummy"

Answer 36

Oh, I see that! Got that point.

Answer 37

Now, how will we find the integrals (after splitting) ?

Answer 38

@thomas5267

we have
\[\int_0^{\infty} \cfrac{x^2+1 }{ 1+x^4}\,dx +a\int_0^{\infty} \cfrac{x }{ 1+x^4}\,dx\]

dividing by \(a\) and taking the limit we see that
\[\begin{align}\lim\limits_{a\to\infty} \frac{1}{a}\left[\int_0^{\infty} \cfrac{x^2+1 }{ 1+x^4}\,dx +a\int_0^{\infty} \cfrac{x }{ 1+x^4}\,dx\right]

&= \lim\limits_{a\to\infty} \frac{1}{a}\left[M+a*N\right]

\end{align}\]

wher \(M\) and \(N\) are the values of definite integrals

Answer 39

Also we know that both integrals converge(by p-series test), so \(M\) and \(N\) are finite real numbers. Therefore the limit, \(\lim\limits_{a\to\infty}\frac{1}{a}M \) equals \(0\).

Answer 40

Need to work the second integral

Answer 41

i would have tried lhopital straight but i see your textbook method isn't bad..

Answer 42

actually it is pretty neat :) that inverse trig identity was applied very nicely!

Answer 43

Fascinating

Answer 44

Thanks rational

Answer 45

rational

how did you go from

\[\int\limits_{0}^{\infty} \frac{ t ^{2}+at+1 }{1+t^{4}}\tan^{-1} t dt\]

to

\[\int\limits_{0}^{\infty} \frac{ x ^{2}+ax+1 }{1+x ^{4}} \tan^{-1} x dx\]

Answer 46

Ahh both give same result as the variable disappears after evaluating definite integral right

http://gyazo.com/57f1203c6dd99f5d8ab65be30a72cfb8

Answer 47

yeah I get that. so then the integration limits must be

\[\int\limits_{\infty}^{0}\]

is that right?

Answer 48

yes i skipped few steps there haha
http://gyazo.com/72aa833609fbce9a98dffeb518026ef5

Answer 49

using this property of definite integrals to swap the bounds

\[\int\limits_a^b f(t) \,dt~~=~~-\int\limits_b^a f(t) \, dt\]

Answer 50

OK, I see. But then why in the blazes do you add the two integrals?
surely this one final integral is sufficient?

Answer 51

that \(\tan^{-1}t\) attached to the integral... is a pain

adding both integrals replaces inverse tangent with a simple number : \(\pi/2\)

Answer 52

I agree it is a pain. but how can you add the two and get away with it?

Answer 53

Ohh ok
again there were manu steps skipped there... let me add them

Answer 54

\(\Large{\color{Red}{\star }} + \Large{\color{blue}{\star }}\) gives

\[\begin{align}I+I&=\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4} . \color{blue}{\tan^{-1} (1/x)}~dx+\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4} . \color{blue}{\tan^{-1} (x)} ~dx\\~\\
&=\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4} . \left(\color{blue}{\tan^{-1} (1/x)+\tan^{-1} (x)}\right)~dx\\~\\
&=\int_0^{\infty} \cfrac{x^2+ax+1 }{ 1+x^4} . \left(\color{blue}{\frac{\pi}{2}}\right)~dx\\~\\
\end{align}\]

Answer 55

below is an identity \[\tan^{-1} (1/x)+\tan^{-1} (x) = \frac{\pi}{2}\]

Answer 56

that's fine, I understand all that and thanks for typing it all up.
But, why add them in the first place? doesn't seem mathematically correct to me?

Answer 57

why, we're allowed to add two equations right :

\[I = \spadesuit\]
\[I=\clubsuit\]

Adding we get
\[I+I =\spadesuit+\clubsuit \]

Answer 58

i see nothing wrong in doing that hmm

Answer 59

so you're adding two integrals in order to solve one integral. I still can't see it?

Answer 60

suppose
\[I=2\]
and
\[I=\sqrt{4}\]

then we have
\[I+I=2+\sqrt{4}\]

Answer 61

yes!

Answer 62

the main idea behind adding is to use that inverse tangent identity and simplify the overall integral

Answer 63

and then you would have to subtract the final answer by π/4 , would that be right?

Answer 64

yes something like that... i did not conclude the final answer haha

Answer 65

i think we need to divide 2 in the end..

Answer 66

we have \(I+I\) on left hand side
which is same as \(2I\)

dividing by \(2\) both sides isolates the \(I\)
just algebra hyeah

Answer 67

I'm with you now! We are just adding two different forms of the same integral, hence divide by 2 for the final answer to the original integral!
Brilliant, thank you.

Answer 68

all done to get rid of that pesky inverse tangent

Answer 69

Exactly! all that circus only to get a simple looking integrand

Answer 70

OK, thanks for going thru that with me. It was bugging me.
I'm signing off for now

Answer 71

np good day/night!