Hmm. It seems like someone says this every time I figure out a nice integral. What was the question?

Are we going to see your working? I'm keen to see if you did it the slick way or the long winded way (or maybe some other way).

Reply 1586

Zacken

22

Original post by atsruser

I'm keen to see if you did it the slick way or the long winded way (or maybe some other way).

Is the slick way

u = \frac{b}{x^n} + a

sub or is there something even slicker? FWIW

Unparseable latex formula:

\displaystyle [br]\begin{align*} \int \frac{\mathrm{d}x}{x^n (a + bx^{1-n}} &= -\frac{1}{b(n-1)} \int \frac{\mathrm{d}u}{u} \\& = -\frac{1}{b(n-1)} \log u \end{align*}

Original post by solC

(where k = a+b, can't seem to get it to show properly with a+b?) which leads to the result

\frac{1}{b(n-1)}ln\left[ \frac{a+b}{a}\right]

.

Write _{a+b}. Anything over two characters long and placed in subsripts/superscripts should be encased in curly brackets, otherwise LaTeX wouldn't know whether you wanted a, a+, or a+b as a subscript if you put in just _a+b.

(edited 6 years ago)

Reply 1587

solC

14

Original post by atsruser

Hmm. It seems like someone says this every time I figure out a nice integral. What was the question?

Are we going to see your working? I'm keen to see if you did it the slick way or the long winded way (or maybe some other way).

It was a special case of your one with a=1, b=-1 (It's from paper III 2005 if you want to see the full question).

This is how I did it:

Write the integral as

\int_{1}^{\infty}\frac{dx}{x(ax^{n-1}+b)}

then let

u = ax^{n-1}+b

giving

dx=\frac{xdu}{(n-1)(u-b)}

after some rearranging. Then we have

\int_{a+b}^{\infty}\frac{dx}{u(u-b)(n-1)} =\left[ln(\frac{u-b}{u}) \right]_{a+b}^\infty

which leads to the result

\frac{1}{b(n-1)}ln\left[ \frac{a+b}{a}\right]

.

Also, for the previous question you posted you said there were 3 ways of approaching it - what was the 3rd way?
I just tried using DUTIS together with induction and it ended up being quite short.

(edited 6 years ago)

Reply 1588

Zacken

22

For the creative A-Levellers,

Unparseable latex formula:

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

Big hint: You may use/assume

\displaystyle \sum_{k\in \mathbb{Z}} \left(k + \frac{1}{n}\right)^{-2} = \pi^2 \csc^2 \left(\frac{\pi}{n} \right).

Reply 1589

username1162972

18

Original post by Zacken

For the creative A-Levellers,

Unparseable latex formula:

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

Big hint: You may use/assume

\displaystyle \sum_{k\in \mathbb{Z}} \left(k + \frac{1}{n}\right)^{-2} = \pi^2 \csc^2 \left(\frac{\pi}{n} \right).

Thanks for letting me assume that xx

Reply 1590

Zacken

22

Original post by physicsmaths

Thanks for letting me assume that xx

lmao

Reply 1591

pcirise

2

Original post by Zacken

lmao

You're a TSR maths legend! RESPECT!!

Reply 1592

atsruser

11

Original post by Zacken

Is the slick way

u = \frac{b}{x^n} + a

sub or is there something even slicker?

Yes, that's the slick way. Did everyone but me know it already? I've never seen it before.

Reply 1593

atsruser

11

Original post by solC

Also, for the previous question you posted you said there were 3 ways of approaching it - what was the 3rd way?
I just tried using DUTIS along with induction and it ended up being quite short.

1. Contour integration
2. Trig substitution + reduction formula, a la Zacken.
3. DUTIS on

\int dx/(x^2+a^2)

and as an added free bonus...

4. there is a reduction formula directly for

\int dx/(x^2+a^2)^n

in terms of rational polynomials.

Reply 1594

DFranklin

18

Original post by atsruser

4. there is a reduction formula directly for

\int dx/(x^2+a^2)^n

in terms of rational polynomials.

For some reason people always seem to find this particular integral difficult. A key fact to realise is that a relation between

I_n

and

I_{n+1}

works perfectly well for reduction, you don't need to start from

I_n

and end up at

I_{n-1}

Spoiler

(edited 6 years ago)

Reply 1595

solC

14

Original post by atsruser

Easyish one for a bedtime treat.

Show that

\displaystyle \int_1^\infty \frac{\cosh x(\ln x+1)-\sinh x}{x^x} \ dx = \cosh 1

Writing everything in terms of exponentials we have

\displaystyle \int_1^\infty \frac{\cosh x(\ln x+1)-\sinh x}{x^x} \ dx = \displaystyle \int_1^\infty \frac{2e^{-x}+lnx(e^x+e^{-x})}{2e^{xlnx}} \ dx

(writing

x^x

as

e^{xlnx}

)

Multiply through by e^x and rearrange to give

\displaystyle \int_1^\infty \frac{(2+lnx)+e^{2x}lnx}{2e^{xlnx+x}} \ dx

Now we can split this into two separate integrals which can both be integrated via recognition.

\displaystyle \int_1^\infty \frac{(2+lnx)}{2e^{xlnx+x}} \ dx + \displaystyle \int_1^\infty \frac{lnx}{2e^{xlnx-x}} \ dx = \frac{-1}{2}\left[ e^{-x(lnx+1))} \right]_1^\infty + \frac{-1}{2}\left[ e^{-x(lnx-1))} \right]_1^\infty = \frac{e^1+e^{-1}}{2}= \cosh1

(edited 6 years ago)

Original post by metrize

Not very experienced with integrals tbh and there's probably an easier way to do this, but after T-Sub, replacing variables with pi/2 - x, two substitutions and transforming to hyperbolics, I get:

\displaystyle I = 8 \pi \int_{-\infty}^{\infty} \frac{x^{4}\cosh{x}}{\cosh{2x}} \mathrm{d}x

which according to Wolfram is numerically the same (I'm an applied mathematician, that's enough for me lol). I then try a rectangular contour from -R to R and R+2*i*pi to -R + 2*i*pi, which does give me

\pi^6 \sqrt{2}

flying about but I get the wrong numbers.

(edited 6 years ago)

Reply 1598

atsruser

11

Original post by solC

Writing everything in terms of exponentials we have
...

Nice! I actually cheated with this. I couldn't be bothered to think too hard, so I went to Wolfram Alpha and differentiated a few functions till I came up with something that looked nice.

And similarly along those lines, show that:

\displaystyle \int_1^\infty \frac{\ln x(\cosh x + \sinh x)}{x^x} \ dx =e

Reply 1599

IrrationalRoot

20

Original post by atsruser

Nice! I actually cheated with this. I couldn't be bothered to think too hard, so I went to Wolfram Alpha and differentiated a few functions till I came up with something that looked nice.

And similarly along those lines, show that: