The hard integral thread.

Given

\displaystyle I_n = \int_{0}^{a}x^{n+0.5}\sqrt{a-x} \mathrm dx

,

n \in \mathbb{Z}

,

n \geqslant 0

Show that

\displaystyle I_n = \Big(\frac{a}{4}\Big)^n \begin{pmatrix}2n+2 \\ n \end{pmatrix}\frac{I_0}{n+1}

,

n \geqslant 1

Determine the value of

\displaystyle \int_{0}^{2}x^{10}\sqrt{4-x^2} \mathrm dx

(edited 6 years ago)

Reply 1681

Zacken

22

Original post by I hate maths

Given

\displaystyle I_n = \int_{0}^{a}x^{n+0.5}\sqrt{a-x} \mathrm dx

,

n \in \mathbb{Z}

,

n \geqslant 0

Show that

\displaystyle I_n = \Big(\frac{a}{4}\Big)^n \begin{pmatrix}2n+2 \\ n \end{pmatrix}\frac{I_0}{n+1}

,

n \geqslant 1

Determine the value of

\displaystyle \int_{0}^{2}x^{10}\sqrt{4-x^2} \mathrm dx

This reads too much like an FP2 question for my liking, nevertheless... :tongue:

Unparseable latex formula:

\displaystyle[br]\begin{align*}3I_n &= 2\left(n+\frac{1}{2}\right)\int_0^a x^{n-0.5}(a-x)\sqrt{a-x} \, \mathrm{d}x \\ &= (2n+1)\left[aI_{n-1} - I_n\right]\end{align*}

via IBP, then we have

Unparseable latex formula:

\displaystyle[br]\begin{align*}I_n &= \frac{2n+1}{2n+4}aI_{n-1} \\ &= \frac{8a^n(2n+1)!!}{(2n+4)!!}I_0 \\ & = \frac{8a^n(2n+2)!}{2^{n+1}(n+1)! \cdot 2^{n+2}(n+2)!} I_0 \\ & = \left(\frac{a}{4}\right)^n {{2n+2} \choose {n}} \frac{I_0}{n+1} \end{align*}

For the last integral, just use

x \mapsto x^2

to get

Unparseable latex formula:

\displaystyle [br]\begin{align*}\int_0^2 x^{10}\sqrt{4-x^2} \, \mathrm{d}x & = \frac{1}{2}\int_0^4 x^{4.5} \sqrt{4-x} \, \mathrm{d}x = \frac{1}{2} {{10} \choose 4} \frac{2\pi}{5} = 42\pi\end{align*}

Reply 1682

username3555092

14

Original post by Zacken

x

Work of a master craftsman... I've learned a lot from reading it like how to align stuff in the middle of the page, here let me try.

Unparseable latex formula:

\displaystyle[br][br]\begin{align*} \int_{1}^2 1 \mathrm{d}x &= 2-1\\ &=1\end{align*}

Omg it worked!!

This is very aesthetically pleasing.

Prove that

\displaystyle \int_0^\infty \frac 1 {xe^x} \left(\frac 1 2 - \frac 1 x + \frac 1 {e^x - 1}\right) \, \mathrm dx = 1 - \frac 1 2 \log 2\pi

(consider expressions of the digamma function)

Reply 1684

ciberyad

8

Original post by Indeterminate

This is a nice one :biggrin:

Evaluate

\displaystyle \int_{0}^{\infty} \dfrac{\ dx}{\sqrt{x}(x^2 + kx + 1)(1-x+x^2-x^3 + \cdots (-x)^n)}

Assuming that

n

is even, this is

I=\displaystyle \int_{0}^{\infty} \dfrac{(1+x)\mathrm dx}{\sqrt{x}(x^2 + kx + 1)(1+x^{n+1})}

(if

n

is odd, the integral does not converge.)

The substitution

x\mapsto x^{-1}

gives

I=\displaystyle \int_{0}^{\infty} \dfrac{(1+x)x^{n+1}\mathrm dx}{\sqrt{x}(x^2 + kx + 1)(1+x^{n+1})}

.

Adding,

2I=\displaystyle \int_{0}^{\infty} \dfrac{(1+x)\mathrm dx}{\sqrt{x}(x^2 + kx + 1)}

.

Consider the substitution

y=\sqrt x-\dfrac1{\sqrt x}

. This gives

\mathrm dy=\dfrac{1+x}{2x\sqrt x}\,\mathrm dx

, so that

y

increases monotonically from

-\infty\textrm{ to }\infty

as

x

goes from

0\textrm{ to }\infty

.

Also

y^2=x+\dfrac1x-2

.

Thus

I=\displaystyle \int_{0}^{\infty} \dfrac{(1+x)\mathrm dx}{2x\sqrt{x}\left(x + k + \dfrac1x\right)}=\int_{-\infty}^\infty \dfrac{\mathrm dy}{y^2+k+2}=\dfrac\pi{\sqrt{k+2}}

.

Reply 1685

ζ(z)

9

Original post by _gcx

This is very aesthetically pleasing.

Prove that

\displaystyle \int_0^\infty \frac 1 {xe^x} \left(\frac 1 2 - \frac 1 x + \frac 1 {e^x - 1}\right) \, \mathrm dx = 1 - \frac 1 2 \log 2\pi

(consider expressions of the digamma function)

LHS Looks ugly but the RHS is pretty.

(edited 5 years ago)

Reply 1686

ciberyad

8

Original post by _gcx

This is very aesthetically pleasing.

Prove that

\displaystyle \int_0^\infty \frac 1 {xe^x} \left(\frac 1 2 - \frac 1 x + \frac 1 {e^x - 1}\right) \, \mathrm dx = 1 - \frac 1 2 \log 2\pi

(consider expressions of the digamma function)

In what follows assume that

t>0

, that all integrals converge, and that all limits taken are valid.

Start with the Abramowitz & Stegun formula for the digamma function

\psi(t)=\dfrac{\Gamma' (t)}{\Gamma (t)}=\displaystyle \int_0^\infty\left[\dfrac{\mathrm e^{-x}}x-\dfrac{\mathrm e^{-tx}}{1-\mathrm e^{-x}}\right]\mathrm dx

.

We also need the well-known identities

\dfrac1t=\displaystyle \int_0^\infty \mathrm e^{-tx}\mathrm dx

and

\ln t=\displaystyle \int_0^\infty \dfrac{\mathrm e^{-x}-\mathrm e^{-tx}}x\mathrm dx

.

Thus

\psi(t+1)-\ln t-\dfrac1{2t}=\displaystyle \int_0^\infty\left[\dfrac{\mathrm e^{-tx}}x-\dfrac{\mathrm e^{-(t+1)x}}{1-\mathrm e^{-x}}-\dfrac{\mathrm e^{-tx}}2\right]\mathrm dx

{}=\displaystyle \int_0^\infty\mathrm e^{-tx}\left[\dfrac1x-\dfrac1{\mathrm e^x-1}-\dfrac12\right]\mathrm dx

. Now, integrating both sides with respect to

t

leads to

\ln\Gamma(t+1)-t\,\ln t+t-\frac12\ln t+c=\displaystyle \int_0^\infty\dfrac{\mathrm e^{-tx}}x\left[\dfrac12-\dfrac1x+\dfrac 1{\mathrm{e}^x-1}\right]\mathrm dx

, where

c

is a constant. (*)

According to Stirling's theorem,

\lim\limits_{t\to\infty}\dfrac{ \Gamma(t+1)}{t^{t+1/2}\mathrm e^{-t}}=\sqrt{2\pi}

, so that

\lim\limits_{t\to\infty}[\,\ln \Gamma(t+1)-(t+\frac12)\ln t+t\,]=\frac12\ln 2\pi

.

Letting

t

tend to infinity in (*) thus gives

\frac12\ln 2\pi+c=0

.

Finally, putting

t=1

in (*) gives

\displaystyle \int_0^\infty \frac 1{x\mathrm e^x} \left(\frac 12 - \frac 1x + \frac 1{\mathrm e^x - 1}\right) \, \mathrm dx = 1 +c=1-\textstyle\frac 12 \ln 2\pi

.

Original post by ciberyad

In what follows assume that

t>0

, that all integrals converge, and that all limits taken are valid.

Start with the Abramowitz & Stegun formula for the digamma function

\psi(t)=\dfrac{\Gamma' (t)}{\Gamma (t)}=\displaystyle \int_0^\infty\left[\dfrac{\mathrm e^{-x}}x-\dfrac{\mathrm e^{-tx}}{1-\mathrm e^{-x}}\right]\mathrm dx

.

We also need the well-known identities

\dfrac1t=\displaystyle \int_0^\infty \mathrm e^{-tx}\mathrm dx

and

\ln t=\displaystyle \int_0^\infty \dfrac{\mathrm e^{-x}-\mathrm e^{-tx}}x\mathrm dx

.

Thus

\psi(t+1)-\ln t-\dfrac1{2t}=\displaystyle \int_0^\infty\left[\dfrac{\mathrm e^{-tx}}x-\dfrac{\mathrm e^{-(t+1)x}}{1-\mathrm e^{-x}}-\dfrac{\mathrm e^{-tx}}2\right]\mathrm dx

{}=\displaystyle \int_0^\infty\mathrm e^{-tx}\left[\dfrac1x-\dfrac1{\mathrm e^x-1}-\dfrac12\right]\mathrm dx

. Now, integrating both sides with respect to

t

leads to

\ln\Gamma(t+1)-t\,\ln t+t-\frac12\ln t+c=\displaystyle \int_0^\infty\dfrac{\mathrm e^{-tx}}x\left[\dfrac12-\dfrac1x+\dfrac 1{\mathrm{e}^x-1}\right]\mathrm dx

, where

c

is a constant. (*)

According to Stirling's theorem,

\lim\limits_{t\to\infty}\dfrac{ \Gamma(t+1)}{t^{t+1/2}\mathrm e^{-t}}=\sqrt{2\pi}

, so that

\lim\limits_{t\to\infty}[\,\ln \Gamma(t+1)-(t+\frac12)\ln t+t\,]=\frac12\ln 2\pi

.

Letting

t

tend to infinity in (*) thus gives

\frac12\ln 2\pi+c=0

.

Finally, putting

t=1

in (*) gives

\displaystyle \int_0^\infty \frac 1{x\mathrm e^x} \left(\frac 12 - \frac 1x + \frac 1{\mathrm e^x - 1}\right) \, \mathrm dx = 1 +c=1-\textstyle\frac 12 \ln 2\pi

.

Brilliant!

Reply 1688

Kummer

7

Ah, this thread was so much fun!

Original post by Kummer

Ah, this thread was so much fun!

The legend himself :redface:

Reply 1690

Zacken

22

Original post by Kummer

Ah, this thread was so much fun!

Good days! You just finished second year at Uni?

Reply 1691

Kummer

7

Original post by _gcx

The legend himself :redface:

I agree, Ernst Kummer is such a legend! :wink:

I mean what's not to like: he wedded algebra and number theory to give us a beautiful baby known as algebraic number theory!

Need to keep this thread alive.

A sum that took me a bit of time to get:

\displaystyle \sum_{n = 1}^\infty \arctan \left({\frac 1 {8n^2}}\right)

(think of expressions for sine)

And of course some fun with Feynman's trick:

\displaystyle \int_0^1 \frac{\arctan x} {x \sqrt{1 - x^2}} \mathrm dx

arctg(x*x)

Original post by _gcx

A sum that took me a bit of time to get:

\displaystyle \sum_{n = 1}^\infty \arctan \left({\frac 1 {8n^2}}\right)

(think of expressions for sine)

Liked this one - when you say sine expression, guess you mean Euler's infinite product if we did it the same way.

Original post by RichE

Liked this one - when you say sine expression, guess you mean Euler's infinite product if we did it the same way.

Yeah same method :smile:

Would be interested if there were an elementary way to evaluate it.

Reply 1696

RichE

15

Original post by _gcx

Yeah same method :smile:

Would be interested if there were an elementary way to evaluate it.

You could probably work out the sum of 8n^2/(x^2+64n^4) using complex analysis, and then integrate your answer.

Reply 1697

ciberyad

8

Original post by _gcx

Need to keep this thread alive.

A sum that took me a bit of time to get:

\displaystyle \sum_{n = 1}^\infty \arctan \left({\frac 1 {8n^2}}\right)

(think of expressions for sine)

Start with the identities

\cot x=\dfrac1x+\displaystyle\sum_{n=1}^\infty \dfrac{2x}{x^2-n^2\pi^2}

, and

\coth x=\dfrac1x+\displaystyle\sum_{n=1}^\infty \dfrac{2x}{x^2+n^2\pi^2}

,

so that

\pi\cot \pi x-\pi\coth \pi x=\displaystyle\sum_{n=1}^\infty \left(\dfrac{2x}{x^2-n^2}-\dfrac{2x}{x^2+n^2} \right)=\sum_{n=1}^\infty \dfrac{4n^2x}{x^4-n^4}

.

Then

\pi\cot \pi(1+\mathrm i)x-\pi\coth \pi(1+\mathrm i)x=-(1+\mathrm i) \displaystyle\sum_{n=1}^\infty \dfrac{4n^2x}{4x^4+n^4}

. (*)

For simplicity, let

u=\tan \pi x

and

v=\tanh \pi x

.

Then

\cot \pi(1+\mathrm i)x-\coth \pi(1+\mathrm i)x=\dfrac{1-\mathrm iuv}{u+\mathrm iv}-\dfrac{1+\mathrm iuv}{v+\mathrm iu}

=(1+\mathrm i)\dfrac{u-v-uv(u+v)}{u^2+v^2}=(1+\mathrm i)\dfrac{u(1-v^2)-v(1+u^2)}{u^2+v^2}

=\left(\dfrac{1+\mathrm i}\pi\right) \dfrac{u\dfrac{\mathrm dv}{\mathrm dx}-v\dfrac{\mathrm du}{\mathrm dx}}{u^2+v^2}=\left( \dfrac{1+\mathrm i}\pi \right) \dfrac{\mathrm d}{\mathrm dx}\arctan \dfrac vu

. (**)

Therefore, from (*) and (**),

\displaystyle\sum_{n=1}^\infty \dfrac{4n^2x}{4x^4+n^4}=-\dfrac{\pi}{1+\mathrm i}(\cot \pi(1+\mathrm i)x-\pi\coth \pi(1+\mathrm i)x)

=-\dfrac{\mathrm d}{\mathrm dx}\arctan \left(\dfrac{\tanh \pi x}{\tan \pi x} \right)

.

Note also that

\dfrac{4n^2x}{n^4+4x^4}= \dfrac{\mathrm d{}}{\mathrm dx} \arctan \left( \dfrac{2x^2}{n^2} \right)

.

Therefore

\dfrac{\mathrm d}{\mathrm dx} \displaystyle \sum_{n=1}^\infty \arctan \left( \dfrac{2x^2}{n^2} \right)=-\dfrac{\mathrm d}{\mathrm dx} \arctan \left( \dfrac{\tanh \pi x}{\tan \pi x} \right)

.

Thus

\displaystyle\sum_{n=1}^\infty \arctan \left( \dfrac{2x^2}{n^2} \right)=-\arctan \left( \dfrac{\tanh \pi x}{\tan \pi x} \right)+C

.

Letting

x\to 0

, we have

0=-\arctan 1+C

, so

C=\dfrac\pi4

.

Thus

\displaystyle\sum_{n=1}^\infty \arctan \left( \dfrac{2x^2}{n^2} \right)=\dfrac\pi4-\arctan \left( \dfrac{\tanh \pi x}{\tan \pi x} \right)

,

which is valid for

0 \leq x \leq 1

(where RHS is continuous.)

Putting

x=\dfrac14

, we get

\displaystyle\sum_{n=1}^\infty \arctan \left(\dfrac1{8n^2}\right)={} \dfrac{\pi}{4}-\arctan \left(\tanh \dfrac\pi4 \right)

.

(edited 5 years ago)

Reply 1698

username4446160

10

Original post by Dabo_26

that was my FP2 prep a few days ago lol

Oof lol
Can't wait for Further Maths AS A-Level next year :biggrin:

The hard integral thread.

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