The hard integral thread.

Original post by Kummer

Calculate

\displaystyle \int_0^1\frac{x(1+x)\sin(\ln(x))}{\ln{x}} ~ dx

I am very late to all this tonight...
I got pi/4
any good?

Reply 62

joostan

13

Original post by TeeEm

I am very late to all this tonight...
I got pi/4
any good?

Agreed.

Reply 63

Original post by joostan

Agreed.

thanks
(mine is unconventional a bit)

Reply 64

username1162972

18

Original post by Krollo

Subbing. Let's integrate :sexface:

From one of the trinity pre interview tests: integrate 1/(x + root(1-x^2))

Posted from TSR Mobile

Let x=sinu or somethingg of the sort.
Then you have an integral cosu/(sinu +cosu). This is when the limits are 1,0.
Then sub of pi/2-n and add them.

Posted from TSR Mobile

Reply 65

Original post by Louisb19

Had fun discussing integrals with you lot yesterday. Lets all post some more and restrict the discussion to this thread.

I'll start

I_n = \displaystyle\int_0^1 (1-\sqrt{x})^n \,dx

Find a reduction formula.

Original post by ghostwalker

I had to cheat, as it's easy enough to integrate directlym via a substitution. Having found the integral, I then construct a reduction formula.

Spoiler

or be completely different

http://www.thestudentroom.co.uk/showthread.php?t=3739401 post 46

Reply 66

Original post by TeeEm

I am very late to all this tonight...
I got pi/4
any good?

Indeed.

Reply 67

Original post by Kummer

Calculate

\displaystyle \int_0^1\frac{x(1+x)\sin(\ln(x))}{\ln{x}} ~ dx

could not resist it ...

there is little appreciation in power of the Laplace transform as an Integration tool.

(it is most of the time DUTIS with standard results, the parameter being s but it can cater for a lot more types of integral than DUTIS)

I may make a thread tomorrow

Original post by TeeEm

could not resist it ...

there is little appreciation in power of the Laplace transform as an Integration tool.

(it is most of the time DUTIS with standard results, the parameter being s but it can cater for a lot more types of integral than DUTIS)

I may make a thread tomorrow

Beautiful solution!

Calculate

\displaystyle \int_0^{\infty} \cot^{-1}{ax}\cot^{-1}{bx}\;{dx}, ~~ a,b > 0

Reply 69

Original post by Kummer

Beautiful solution!

Calculate

\displaystyle \int_0^{\infty} \cot^{-1}{ax}\cot^{-1}{bx}\;{dx}, ~~ a,b > 0

job for tomorrow ... If I get a minute.

(rearrange the integrand into separate arctans perhaps)

Reply 70

Original post by Kummer

Calculate

\displaystyle \int_0^1\frac{x(1+x)\sin(\ln(x))}{\ln{x}} ~ dx

\displaystyle \int_0^1\frac{x(1+x)\sin(\ln(x))}{\ln{x}} ~ dx = \int_0^1 \int_0^1 x(1+x)\cos(t\ln{x})\;{dt}\;{dx} = \\ \int_0^1 \int_0^1 x(1+x)\cos(t\ln{x})\;{dx}\;{dt} = \int_0^1 \frac{3}{t^2+9}+\frac{2}{t^2+4} \;{dt} \\ = \tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\bigg(\frac{\frac{1}{3}+ \frac{1}{2}}{1- \frac{1}{2}\cdot \frac{1}{3}}\bigg) = \tan^{-1}{1} = \frac{\pi}{4}.[br]

Reply 71

Gregorius

14

Original post by Krollo

Is there a way to determine this integral exactly, out of interest?

If we do the "obvious" substitutions in the integral

\displaystyle I = \int_{\frac{\pi^2}{16}}^{\frac{ \pi ^2}{4}} \frac{\sin \sqrt{x}}{x} \ dx

then we soon find that the answer is

\displaystyle 2 Si( \pi / 2) - 2 Si( \pi / 4)

where Si is the "Sine Integral" special function

\displaystyle Si(x) = \int_{0}^{x} \frac{\sin t}{t} \ dt

Unlikely to have an expression in terms of more elementary functions.

Reply 72

DFranklin

18

Original post by joostan

A little messy, but ya'll might like it:

\displaystyle\int \dfrac{1}{1+x^4} \ dx

Obviously a classic. If you want to have the same challenges with a little less mess, you can instead look at

\displaystyle \int \dfrac{1}{x^4+4} \, dx

Spoiler

Reply 73

Original post by Kummer

Beautiful solution!

Calculate

\displaystyle \int_0^{\infty} \cot^{-1}{ax}\cot^{-1}{bx}\;{dx}, ~~ a,b > 0

I tried this for around an hour
It has "ate" me, chewed me and spat me out; plain and simple.

(I could cheat and search the serious forums but I chose not to just in case I look at it again)

Reply 74

Original post by TeeEm

I tried this for around an hour
It has "ate" me, chewed me and spat me out; plain and simple.

(I could cheat and search the serious forums but I chose not to just in case I look at it again)

It's a tough one!

Spoiler

Reply 75

Calculate

\displaystyle \int_0^\infty {\left\lfloor {\log _2 \left\lfloor {\frac{{\left\lceil x \right\rceil }}{x}} \right\rfloor } \right\rfloor \,dx}

Come on, people! I'm sure you all have some nice integrals to share.

Reply 76

Original post by Kummer

It's a tough one!

Spoiler

I have done it essentially (3/4 of the way and from there I can finish it)
This was very long too, but definitely worth it
(this is definitely "undergrad STEP" level)

Reply 77

math42

20

Original post by Kummer

Calculate

\displaystyle \int_0^\infty {\left\lfloor {\log _2 \left\lfloor {\frac{{\left\lceil x \right\rceil }}{x}} \right\rfloor } \right\rfloor \,dx}

Come on, people! I'm sure you all have some nice integrals to share.

Is it 1...

Spoiler

Reply 78

Original post by 1 8 13 20 42

Is it 1...

Spoiler

Yes. In fact, for all a > 0 we have

\displaystyle \int_0^\infty {\left\lfloor {\log _a \left\lfloor {\frac{{\left\lceil x \right\rceil }}{x}} \right\rfloor } \right\rfloor \,dx}= ~ ?

Spoiler

Reply 79