The Student Room Group

Integral help

Any good method for this?
(please attach I do not have home-works any more!!!)

I have the correct answer (which is part of an undergrad question to which I am writing a solution)
It is just that the method seems a bit too long and I do not want to rewrite the solution as this integral lies somewhere in the middle of a longish problem

Thanks
(edited 8 years ago)

Scroll to see replies

TeeEm needing help?!? Thought I'd never see the day! :P
Reply 2
Integration by parts?
Reply 3
Original post by iMacJack
TeeEm needing help?!? Thought I'd never see the day! :P


happens to me more regularly than I would like to admit ...

However in this question I have done the integral without problems; I just look for a quicker alternative if it exists.
( it is just that I am doing high level maths non-stop since this morning and my brain is now stalling )
Original post by ubisoft
Integration by parts?


I second that idea...and so does Wolfram Alpha 🙂
Original post by TeeEm
happens to me more regularly than I would like to admit ...

However in this question I have done the integral without problems; I just look for a quicker alternative if it exists.
( it is just that I am doing high level maths non-stop since this morning and my brain is now stalling )


Lol! Fair enough man. My brain gets fried from too much FP1, so you're not doing too badly :P Keep up the good work man
Reply 6
Original post by Gregorius
I second that idea...and so does Wolfram Alpha 🙂


I split the log

the second part I left alone

The first part of the log
I did substitution x = 2tantheta for the first part of the log
then parts

then combined all into the answer


EDIT No need for substitution/ direct parts cuts the work a bit/ good enough
(edited 8 years ago)
Original post by TeeEm
I split the log


Mathematical lumberjack as well now!
Reply 8
Original post by Gregorius
Mathematical lumberjack as well now!


I cut a bit of the work out now, so it is ok...

It is actually one of the questions you posed the other night, that I took and made a question out of ...

Find

INTEGRAL (sin2x/x2) from 0 to infinity, using Laplace transform techniques
Reply 9
Original post by Gregorius
Mathematical lumberjack as well now!


Original post by TeeEm
I cut a bit of the work out now, so it is ok...

It is actually one of the questions you posed the other night, that I took and made a question out of ...

Find

INTEGRAL (sin2x/x2) from 0 to infinity, using Laplace transform techniques


This reminds me, Greogrius - I tried computing a recurrence for the integral of nth powers of the since function, i.e a closed form for 0(sinxx)ndx\displaystyle \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x and eventually gave up (although I managed to get the first few nth powers, up to 5 or so) - then a quick math.stackexchange lookup yielded and impressive and complex-looking closed form!
Reply 10
Original post by Zacken
This reminds me, Greogrius - I tried computing a recurrence for the integral of nth powers of the since function, i.e a closed form for 0(sinxx)ndx\displaystyle \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x and eventually gave up (although I managed to get the first few nth powers, up to 5 or so) - then a quick math.stackexchange lookup yielded and impressive and complex-looking closed form!


I just borrowed the n=2 case to make a question for my LaplaceTransform resources
Reply 11
Original post by TeeEm
I just borrowed the n=2 case to make a question for my LaplaceTransform resources


I saw - I really need to brush up on my Laplace Transforms, perhaps starting with your resources! :-)
Original post by Zacken
This reminds me, Greogrius - I tried computing a recurrence for the integral of nth powers of the since function, i.e a closed form for 0(sinxx)ndx\displaystyle \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x and eventually gave up (although I managed to get the first few nth powers, up to 5 or so) - then a quick math.stackexchange lookup yielded and impressive and complex-looking closed form!


Ooh! Do you have a link to the math.stackexchange article?
Reply 13
Original post by Gregorius
Ooh! Do you have a link to the math.stackexchange article?


Here you go, although most of the information/maths is in the PDFs/papers that they've linked to in the comments. :-)
Original post by TeeEm
Any good method for this?
(please attach I do not have home-works any more!!!)

I have the correct answer (which is part of an undergrad question to which I am writing a solution)
It is just that the method seems a bit too long and I do not want to rewrite the solution as this integral lies somewhere in the middle of a longish problem

Thanks


I=log(x2+1x2) dxI=\int \log(\frac{x^2+1}{x^2}) \ dx

Let x=tanudx=sec2udux=\tan u \Rightarrow dx=\sec^2 u du

So:

I=log(sec2utan2u) du=log(1sin2u)sec2u du=2log(sinu)sec2u duI = \int \log(\frac{\sec^2 u}{\tan^2 u}) \ du = \int \log(\frac{1}{\sin^2 u}) \sec^2 u \ du = -2 \int \log(\sin u) \sec^2 u \ du

Now by parts we have:

I=2{tanulog(sinu)tanucotu du}=2{tanulog(sinu)u}+CI= -2 \{ \tan u \log(\sin u) - \int \tan u \cot u \ du \} = -2 \{ \tan u \log(\sin u) - u \} + C

Now x=tanusinu=xx2+1x=\tan u \Rightarrow \sin u = \frac{x}{\sqrt{x^2+1}} so:

I=2{xlog(xx2+1)tan1x}+C=xlog(x2+1x2)+2tan1x+CI = -2 \{ x \log(\frac{x}{\sqrt{x^2+1}}) - \tan^{-1} x \} + C = x \log(\frac{x^2+1}{x^2}) + 2 \tan^{-1} x + C

(Confirmed by Wolfram)
(edited 8 years ago)
Reply 15
Original post by Zacken
I saw - I really need to brush up on my Laplace Transforms, perhaps starting with your resources! :-)


I wrote many questions last week as I am helping 2 students at present,

wait until I update the site as none of them are there

here is the one I made tonight which lead to this integral
Original post by Zacken
Here you go, although most of the information/maths is in the PDFs/papers that they've linked to in the comments. :-)


Excellent! An endless stream of exam questions for the future!
Reply 17
Original post by Gregorius
Excellent! An endless stream of exam questions for the future!


Poor students. :wink:
Reply 18
Original post by TeeEm
I wrote many questions last week as I am helping 2 students at present,

wait until I update the site as none of them are there

here is the one I made tonight which lead to this integral


Thanks for this! :-)
Reply 19
Original post by atsruser
I=log(x2+1x2) dxI=\int \log(\frac{x^2+1}{x^2}) \ dx

Let x=tanudx=sec2udux=\tan u \Rightarrow dx=\sec^2 u du

So:

I=log(sec2utan2u) du=log(1sin2u)sec2u du=2log(sinu)sec2u duI = \int \log(\frac{\sec^2 u}{\tan^2 u}) \ du = \int \log(\frac{1}{\sin^2 u}) \sec^2 u \ du = -2 \int \log(\sin u) \sec^2 u \ du

Now by parts we have:

I=2{tanulog(sinu)tanucotu du}=2{tanulog(sinu)u}+CI= -2 \{ \tan u \log(\sin u) - \int \tan u \cot u \ du \} = -2 \{ \tan u \log(\sin u) - u \} + C

Now x=tanusinu=xx2+1x=\tan u \Rightarrow \sin u = \frac{x}{\sqrt{x^2+1}} so:

I=2{xlog(xx2+1)tan1x}+C=xlog(x2+1x2)+2tan1x+CI = -2 \{ x \log(\frac{x}{\sqrt{x^2+1}}) - \tan^{-1} x \} + C = x \log(\frac{x^2+1}{x^2}) + 2 \tan^{-1} x + C

(Confirmed by Wolfram)


Thank you

I originally did that way but I did it directly by parts ( a tiny bit quicker)
I was just wondering if there was an obvious method (as I am very tired) and this was part of a solution so I did not want to redo it at a later stage.

Quick Reply

Latest