The hard integral thread.

Original post by Zacken

...

Original post by Farhan.Hanif93

...

Yep, both of these are spot on. Nice solutions.

Reply 1182

Farhan.Hanif93

Original post by ThatPerson

Evaluate

I= \displaystyle \int^{\frac{\pi}{2}}_{0} \arctan{\left(\cos^2{x}\right)}\ \mathrm{d}x

Since a solution hasn't been posted yet, observe that

\arctan(\cos ^2x) = \cos^2x \displaystyle\int_0^1 \dfrac{1}{1+y^2\cos^4 x } dy

so:

Unparseable latex formula:

\begin{array}{lcl}[br]I &=& \displaystyle\int_0^{\pi/2} \displaystyle\int_0^{1} \dfrac{\cos ^2x}{1+y^2\cos ^4x } dydx\\[br]\\[br]&=& \displaystyle\int_0^{1} \displaystyle\int_0^{\pi /2} \dfrac{\sec ^2x}{(1+\tan^2 x)^2 + y^2} dxdy \ \ \ \ \ \ \ \ \ \text{(Exchanging the order of integration)}\\[br]\\[br]&=&\displaystyle\int_0^{1} \displaystyle\int_0^{\infty} \dfrac{1}{(1+t^2)^2 + y^2} dtdy \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Note: } x^2+y^2 = (x+iy)(x-iy))\\ [br]\\ [br]&=&\displaystyle\int_0^{1} \dfrac{1}{2yi}\displaystyle\int_0^{\infty} \left[ \dfrac{1}{(\sqrt{1-iy})^2+t^2} - \dfrac{1}{(\sqrt{1+iy})^2 +t^2}\right] dtdy \\[br]\\[br]&=&\displaystyle\int_0^1 \dfrac{1}{2yi} \left[\dfrac{1}{\sqrt{1-iy}}\arctan\left( \dfrac{t}{\sqrt{1-iy}} \right) - \dfrac{1}{\sqrt{1+iy}} \arctan \left(\dfrac{t}{\sqrt{1+iy}} \right)\right]^{\infty}_0dy\\[br]\\[br]&=&\dfrac{\pi}{2} \displaystyle\int_0^1 \dfrac{1}{2yi} \left(\dfrac{1}{\sqrt{1-iy}} - \dfrac{1}{\sqrt{1+iy}} \right) dy\\[br]\\[br]&=&\dfrac{\pi}{2} \left(\displaystyle\int_0^1 \dfrac{\mathrm{d}(\sqrt{1+iy})}{y} + \displaystyle\int_0^1 \dfrac{\mathrm{d}(\sqrt{1-iy})}{y}\right) \\ [br]\end{array}

Unparseable latex formula:

\begin{array}{lclr} [br]\ \ &=& \dfrac{\pi}{2}\left( \displaystyle \int_1^{\sqrt{1+i}} \dfrac{idz}{z^2-1} - \displaystyle\int_1^{\sqrt{1-i}} \dfrac{idz}{z^2-1} \right) \\[br]\\[br]&=& \dfrac{\pi}{2} \displaystyle\int_{\sqrt{1-i}}^{\sqrt{1+i}} \dfrac{i}{2} \left(\dfrac{1}{z-1} - \dfrac{1}{z+1}\right)dz \\[br]\\[br]&=& \dfrac{\pi}{2} \cdot \dfrac{i}{2} \left[\log\dfrac{z-1}{z+1} \right]_{\bar{\alpha}}^{\alpha} & \text{(Where } \alpha=\sqrt{1+i}=2^{1/4}e^{i\pi /8}\text{)} \\ [br]\\[br]&=& \dfrac{\pi}{2} \cdot \dfrac{i}{2} \log \dfrac{(\alpha-1)(\overline{\alpha}+1)}{(\alpha+1)(\overline{\alpha}-1)}\\[br]\\[br]&=& \dfrac{\pi}{2} \cdot \dfrac{i}{2} \log \dfrac{1+\frac{|\alpha|^2-1}{\alpha - \overline{\alpha}}}{1-\frac{|\alpha|^2-1}{\alpha-\overline{\alpha}}}\\[br]\\[br]&=& \dfrac{\pi}{2} \cdot \dfrac{i}{2} \log \dfrac{1-i\frac{\sqrt 2 - 1}{2\mathrm{Im} \alpha}}{1+i\frac{\sqrt 2 -1}{2\mathrm{Im} \alpha}} [br]\end{array}

Also, observe that:

\mathrm{Im} \alpha = 2^{1/4} \sin (\pi/8) = \sqrt{\sqrt 2} \cdot \dfrac{\sqrt{2-\sqrt 2}}{2} =\sqrt{\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}}

\iff \dfrac{\sqrt{2}-1}{2\mathrm{Im}\alpha} = \dfrac{\frac{1}{\sqrt 2} - \frac{1}{2}}{\mathrm{Im} \alpha} = \sqrt{\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}}

Hence, using the identity

\arctan(z)=\dfrac{i}{2}\log \dfrac{1-iz}{1+iz}

, it follows that:

\boxed{\displaystyle\int_0^{\pi/2} \arctan(\cos ^2 x) dx = \dfrac{\pi}{2} \arctan \sqrt{\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}}}

[Please tell me there is a better way...]

(edited 7 years ago)

Reply 1183

Original post by Zacken

Using IBP with

u=x

and

\mathrm{d}v = \text{tanh}(\alpha x)\text{sech} (\alpha x)

we get:

...

Nice problem, though. Thanks.

OK. That turned out to be a lot easier than I had anticipated. I got to it by showing that

\int_0^\infty \text{sech } \alpha x \ dx = \frac{\pi}{2\alpha}

, and then applying a little light DUTIS to both sides. I didn't notice the easy application of IBP to solve it, I must say.

I do notice, however, that a couple of my earlier problems are still unsolved, to wit:

http://www.thestudentroom.co.uk/showpost.php?p=62139773&postcount=779
http://www.thestudentroom.co.uk/showpost.php?p=62255567&postcount=828

Reply 1184

Original post by Farhan.Hanif93

IIRC I attempted to differentiate w.r.t.

r

, consider a difference for two different values of

r

and manipulated the integral itself to no avail [bar practically doing the integral, which defeats the purpose a little]. Haven't tried again since, however.

From the POV of view of complex methods, it is "obvious" that the integral can't depend upon

r

, since the value of the integral of a holomorphic function at a point depends only on its values on any closed contour around the point - so the circular contour can be as big as we like. However, that doesn't give me much insight into why the real integral should not depend upon

r

- I'd like a non-complex explanation for that, though maybe one doesn't exist, of course.

And, my apologies! I must have missed the quote. There wasn't anything too special/neat - I was just curious about how deep these conditions would stretch (and 'not very far' seems to be the answer).
*

Well, what are the conditions?

Reply 1185

Original post by atsruser

OK. That turned out to be a lot easier than I had anticipated. I got to it by showing that

\int_0^\infty \text{sech } \alpha x \ dx = \frac{\pi}{2\alpha}

, and then applying a little light DUTIS to both sides. I didn't notice the easy application of IBP to solve it, I must say.

I see, I did think of trying DUTIS but then stopped myself when I recognised that

\text{sech} \, \text{tanh}

was a derivative of a hyperbolic trig function.

Your other two problems look hard. :lol:

Reply 1186

I like this one; show that:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_0^{\infty} \sin x \arctan \left(\frac{1}{x}\right) \, \mathrm{d}x = \frac{\pi}{2}\left(1-\frac{1}{e}\right)\end{equation*}

Reply 1187

Original post by Zacken

I see, I did think of trying DUTIS but then stopped myself when I recognised that

\text{sech} \, \text{tanh}

was a derivative of a hyperbolic trig function.

Your other two problems look hard. :lol:

Well, perhaps they will go some little way to make up for the embarrassingly easy question that I just posted, which has brought shame upon my family name, and for which my forefathers will never forgive me.

One of the two is quite complex, I seem to recall.

Reply 1188

Imperion

17

Original post by ThatPerson

Let

\displaystyle I = \int^{\infty}_{0} \frac{\ln x}{x^2 + 2x + 2}\ \mathrm{d}x

. Then

\displaystyle I \stackrel{x=u-1}{=} \int^{\infty}_{1} \frac{\ln \left(u-1\right)}{u^2 + 1} \mathrm{d}u \stackrel{u=\tan\theta}{=} \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \ln \left(\tan\theta - 1\right) \mathrm{d} \theta

Now substitute

\theta \mapsto \frac{3\pi}{4} - \theta

. Notice that

\displaystyle \tan\left( \frac{3\pi}{4} - \theta \right)-1 = -\tan\left(\frac{\pi}{4}+\theta \right) - 1 = \frac{2}{\tan\theta - 1}

.

Hence

\displaystyle I = \ln 2 \left(\frac{\pi}{2}-\frac{\pi}{4}\right) - I \Rightarrow I = \frac{\pi}{8}\ln 2 \ \square

Spoiler

Reply 1189

Lord of the Flies

19

Original post by Zacken

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_0^{\infty} \sin x \arctan \left(\frac{1}{x}\right) \, \mathrm{d}x = \frac{\pi}{2}\left(1-\frac{1}{e}\right)\end{equation*}

Have

f(q)=\displaystyle\int_0^{\infty} \sin x \arctan q x^{-1}\, \mathrm{d}x

. By parts twice:

f(q)=\frac{\pi}{2}+f''(q)

.

The solution to the diff. eq. is

f(q)=\frac{\pi}{2}+\alpha e^{-q}+\beta e^q

. Applying

f(0)=0

and

f'(0)=\frac{\pi}{2}

(classic sinc integral) you get the result.

Reply 1190

Original post by Lord of the Flies

You get the result.

Splendid! I've never thought about thinking of them as differential equations before. I used DUTIS in the first place, had an integral where I needed to apply DUTIS to in order to evaluate it before continuing with the usual stuff, much messier than this. (also makes the seemingly random appearance of

e

more understandable).

Reply 1191

ThatPerson

Original post by Farhan.Hanif93

[Please tell me there is a better way...]

Hehe, there is indeed an easier route. I should say that this solution isn't due to me - I saw it on Stackexchange, though I can't find the original post at the moment.

Picking up from where Zacken left off: We have

\displaystyle f'(\alpha)=\int^{\infty}_{0} \frac{1}{t^4+2t^2+\alpha^{2}+1} \mathrm{d}t

. Now substitute

t \mapsto \frac{\sqrt{\alpha^{2}+1}}{t}

so

\displaystyle f'(\alpha) = \frac{1}{\sqrt{\alpha^{2}+1}} \int^{\infty}_{0} \frac{t^2}{t^4+2t^2+\alpha^{2}+1} \mathrm{d}t

.

Thus by adding the integrals together

\displaystyle f'(\alpha) = \frac{1}{2\sqrt{\alpha^{2}+1}} \int^{\infty}_{0} \frac{\sqrt{1+\alpha^{2}} + t^2}{t^4+2t^2+\alpha^{2}+1} \mathrm{d}t

We can rewrite the integral as

Unparseable latex formula:

\displaystyle \frac{1}{2\sqrt{\alpha^{2}+1}} \int^{\infty}_{0} \frac{\frac{\sqrt{\alpha^{2}+1}}{t^2} + 1}{\left(t-\frac{\sqrt{\alpha^{2}+1}}{t}} \right)^{2} + 2(1+\sqrt{\alpha^{2}+1})}

Let

y = t - \frac{\sqrt{\alpha^{2}+1}}{t}

which results in an

\arctan

integral. We get

\displaystyle f'(\alpha) = \frac{\pi}{2\sqrt{2}} \frac{1}{\sqrt{1+\alpha^{2}} \sqrt{1+\sqrt{1+\alpha^{2}}}}

Recovering

f(\alpha)

is simple.

On another note, I always get an error when I try to use align in LaTeX; can someone give me a quick example on how to use it?

(edited 7 years ago)

Reply 1192

Original post by ThatPerson

On another note, I always get an error when I try to use align in LaTeX; can someone give me a quick example on how to use it?

I'd never have thought of that. Nice solution though. TSR is a bit quirky in the way you use align, you need to indent it in a specific way, like so:

[noparse]
Code block here:

Unparseable latex formula:

\displaystyle[br]\begin{align*} a &= b \\ & = c \\ & = d \end{align*}

[/noparse]

to produce:

Unparseable latex formula:

\displaystyle [br]\begin{align*} a & =b \\ & = c \\ & = d\end{align*}

Note that the \displaystyle needs to be on the same line as the tex tag and the \begin{align*} needs to be in the line below \displaystyle. There shouldn't be a new line before \end{align} either and (I think) the close tex tag needs to be just after the \end{align*}.

Reply 1193

ThatPerson

Original post by Zacken

...

I see; thanks for the example.

Reply 1194

Evaluate:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_0^{\frac{ \pi}{2}} \ln (\alpha^2 \cos^2 x + \sin^2 x) \, \mathrm{d}x \end{equation*}

Reply 1195

Original post by Zacken

Evaluate:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_0^{\frac{ \pi}{2}} \ln (\alpha^2 \cos^2 x + \sin^2 x) \, \mathrm{d}x \end{equation*}

This was a rush job but is it

Spoiler

Reply 1196

Original post by atsruser

This was a rush job but is it

Spoiler

Yep!

Reply 1197

Original post by Zacken

Evaluate:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_0^{\frac{ \pi}{2}} \ln (\alpha^2 \cos^2 x + \sin^2 x) \, \mathrm{d}x \end{equation*}

Spoiler

Reply 1198

Original post by atsruser

...

Spot on!

I did the J integral part like this:

Unparseable latex formula:

\displaystyle [br]\begin{align*}f'(\alpha) &= 2\alpha \int_0^{\pi/2}\frac{\mathrm{d}x}{\alpha^2 + \tan^2 x } \\ & = 2\alpha \int_0^{\pi/2} \frac{\sec^2 x - \tan^2 x}{\alpha^2 + \tan^2 x} \, \mathrm{d}x \\ & = 2\alpha\bigg[\frac{1}{\alpha} \cdot \arctan \frac{x}{\alpha}\bigg]_0^{\infty} - 2\alpha \int_0^{\pi/2} \frac{\tan^2 x}{\alpha^2 + \tan^2 x} \, \mathrm{d}x \\& = \pi - 2\alpha \int_0^{\pi/2} \frac{\alpha^2 + \tan^2 x - \alpha^2}{\alpha^2 + \tan^2 x} \, \mathrm{d}x \\& = \pi - 2\alpha \cdot \frac{\pi}{2} + 2\alpha \int_0^{\pi/2} \frac{\alpha^2}{\alpha^2 + \tan^2 x} \, \mathrm{d}x \end{align*}

That is:

Unparseable latex formula:

\displaystyle [br]\begin{align*}f'(\alpha) &= \pi - \alpha \pi + \alpha^2 f'(\alpha) \\ \Rightarrow f'(\alpha) (1-\alpha^2) &= \pi(1-\alpha) \\ \Rightarrow f'(\alpha) &= \frac{\pi}{1+\alpha} \end{align*}

(edited 7 years ago)

Reply 1199

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