- freckles

Integral Jungle! Are there any integrals you know of that seem like really hard to do but you just need to find a cute way to manipulate things? Oh and also please none that involve non-elementary functions. Thank you kindly.

- katieb

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- freckles

Could be indefinite or definite! Whateves!

- Kainui

Here's a good one if you've never seen this before: \[\large \int\limits_{-\infty}^\infty e^{-x^2}dx\] I know a handful of other fun integrals too that I really enjoy.

- freckles

I think that one is a famous one. I think I remember seeing it done by taylor series.

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## More answers

- freckles

\[e^x=\sum_{n=0}^\infty \frac{x^n}{n!} \\ e^{-x^2}=\sum_{n=0}^\infty \frac{(-x^2)^n}{n!}=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{n!}\]
\[\int\limits_0^\infty e^{-x^2} dx \\ e^{-x^2} \approx 1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+ \cdots\]
hmm maybe my memory isn't right

- freckles

or maybe I just need to play with that idea more

- freckles

nevermind don't think it possible with taylor or power whatever you call it

- Kainui

Haha yeah there's a nice way of doing it. =)

- Kainui

The trick is sorta weird if you've never seen it before. The first step is to square the integral, but you need to rewrite the other integral in terms of a different dummy variable. Since it's squared, you take the square root, so these integrals are exactly the same but the one on the right lets us play around and combine them into a double integral:
\[\large \int\limits_{-\infty}^\infty e^{-x^2}dx = \sqrt{\int\limits_{-\infty}^\infty e^{-x^2}dx \int\limits_{-\infty}^\infty e^{-y^2}dy} \]

- freckles

\[I=\int\limits_0^\infty e^{-x^2} dx \\ I \cdot I=\int\limits_0^\infty e^{-x^2} dx \cdot \int\limits_0^\infty e^{-x^2} dx \\ \text{ change up the variable on one } \\ I^2=\int\limits_0^\infty e^{-x^2} dx \cdot \int\limits_0^\infty e^{-y^2} dy \\ I^2=\int\limits_0^\infty \int\limits_0^\infty e^{-x^2} e^{-y^2} dx dy\]
I don't know why the last two lines are equivalent?

- Kainui

The reason we can combine them together is because they're summed over a different independent variables so the multiplication basically is just distributing.
For example: \[\large \left( \sum_{n=1}^3 a_n \right)\left( \sum_{m=1}^3 b_m \right) = \sum_{n=1}^3 \sum_{m=1}^3 a_n b_m \]
\[(a_1+a_2+a_3)(b_1+b_2+b_3) = a_1b_1+a_1b_2+a_1b_3+ \cdots\]
I don't know if that's convincing enough or not.

- freckles

yes that is actually

- freckles

\[I^2 =\int\limits_0^\infty \int\limits_0^\infty e^{-(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits_0^\infty \int\limits_0^\infty r e^{-r^2} dx dy \\ 0

- Kainui

Awesome, alright I guess I need to find something harder to send your way.

- freckles

lol
when you say harder I know it will be
because you are crazy good at math

- freckles

by the way I did make a mistake in one of my lines

- freckles

if I was going to leave the dx dy there while I figured out the limits
then that one r shouldn't haven been there yet
since dx dy=r dr d theta

- freckles

\[I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{-(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{-r^2} dx dy \\ 0

- Kainui

Hmmm well you seem to have messed up twice or something along the lines so it cancelled out or something?
http://www.wolframalpha.com/input/?i=integral%20-infty%20to%20infty%20e%5E-x%5E2dx&t=crmtb01

- freckles

yeah that is the same I got
I'm just saying I put the r in too early

- Kainui

Ohhh ok I see. Alright here's a fun one haha: \[\large \int\limits_0^\infty \frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx \text{ when } a>0\]

- freckles

now this one I do believe is entirely new to me
let me think a bit

- freckles

so that means a can be 1
and if a is 1 then the integrand is 0
which mean the output is 0 for a=1
so should the inequality include 1?

- freckles

you know what nevermind
it is probably some kinda pattern or whatever

- Kainui

Yep, you might have to be careful about what a is when you're defining your answer, like specify some things perhaps. I don't have the answer with me but it's not too difficult if you know the trick =P
Here's another good one: \[\large \int\limits_0^\infty \lfloor x \rfloor e^{-x} dx\]

- freckles

hey @xapproachesinfinity we are just having integral fun if you want to join in in either trying to answer or give more questions

- Kainui

Here's one that took several days to figure out, but we did it:
Integrate:
\[\int_0^\infty \frac{(\sin x)^{2k+1}}{x}\,dx\]
for non-negative integer \(k\).
This one originally comes from here: http://openstudy.com/study#/updates/54bc2960e4b0c3c33929e555

- xapproachesinfinity

integral party lol
i have seen that first integral many times ant its value never question how it is root(pi)

- freckles

@Kainui is there any more ways to integrate the first one you mentioned besides the user of "polar-ness"
That might be easiest way right?
Like the taylor/power series is a complete dead end?

- freckles

and wow you guys did put a lot of work into that one problem

- freckles

http://openstudy.com/users/sithsandgiggles#/updates/54a1af22e4b054f0c3b34fcf
I never understood this parameter thing

- anonymous

Hi, please accept this monstrosity: http://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1

- Kainui

There's one integral that's really fun to do with integration by parts, but you can also do it with linear algebra, this one:
\[\large \int\limits e^{ax}\cos(bx)dx\] The trick is pretty fun and you can use a similar trick for solving some differential equations too:
Find basis vectors: \[y_1 = e^{ax}\sin(bx) \\ y_2 = e^{ax}\cos(bx)\] The derivative of these functions is a linear combination of these functions so now instead of doing integration by parts, integration is just inverting a 2x2 matrix!

- freckles

like
why does
\[I(b)=\int\limits_0^\infty \frac{\ln(bx)}{1+x^2} dx =\int\limits_0^\infty \frac{\ln(b)+\ln(x)}{1+x^2} dx \\ =\int\limits_0^\infty \frac{\ln(b)}{1+x^2} dx+\int\limits_0^\infty \frac{\ln(x)}{1+x^2} dx \\ =\ln(b) \int\limits_0^\infty \frac{1}{1+x^2} dx+\int\limits_0^\infty \frac{\ln(x)}{1+x^2} dx \\ I'(b)=\frac{1}{b} \int\limits_0^\infty \frac{1}{1+x^2}dx+0\]
oops nevermind I get it

- freckles

@sithsandgiggles has asked a lot of integral questions

- freckles

hard ones too

- freckles

\[\large \int\limits\limits_0^\infty \frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx \text{ when } a>0 \\ f(t)=\tan^{-1}(tx) \\ f(a)=\tan^{-1}(ax) \\ f(1)=\tan^{-1}(1x)=\tan^{-1}(x) \\ \int\limits_0^\infty \frac{f(a)-f(1)}{x} dx \\ \int\limits_0^\infty \frac{1}{x} \int\limits_1^a f'(t) dt\]
trying to play with this

- freckles

though I don't think that is exactly what I want

- xapproachesinfinity

just one thing the first one
how likely a person would think they need to change the dummy variable
isn't that somewhat unlikely?

- freckles

guess it kind of depends on if one is familiar with polar-need integration
if that is the case the odds that person does that goes a little up

- freckles

polar-ness*

- freckles

you know like since we know x^2+y^2=r^2

- xapproachesinfinity

i agree

- freckles

then eventually using the r dr dtheta= dx dy thing
so the integration actually becomes doable

- freckles

\[\int\limits f'(tx) dt \\ \text{ Let } u(t)=tx \\ \frac{d(u(t))}{dt}=\frac{d(tx)}{dt} \\ \frac{d(u(t))}{dt}=x \\ \frac{du}{dt}=x \\ du=x dt \\ \frac{1}{x} du =dt \\ \int\limits f'(tx) dx=\frac{1}{x} \int\limits f'(u) du=\frac{1}{x}( f(u)+C)\]
so maybe I can redefined by f from above
so we have
\[f'(tx)=\frac{\arctan(tx)}{x} \\ f'(ax)=\frac{\arctan(ax)}{x} \\ f'(1x)=\frac{\arctan(1x)}{x} \\ f'(ax)-f'(1x)=\frac{\arctan(ax)}{x}-\frac{\arctan(1x)}{x}=\frac{\arctan(ax)-\arctan(x)}{x} \\ \\ \text{ so we could say } \\ \int\limits_0^\infty \frac{\arctan(ax)-\arctan(x)}{x} dx \\ \int\limits_0^\infty \int\limits_1^a f'(tx) dt dx\]
I don't know if this helps

- freckles

I guess we can try the fubini theorem thingy

- freckles

\[\int\limits_1^a \int\limits_0^\infty f'(tx) dx dt \\ \text{ Let } u(x)=tx \\ u'(x)=t \\ du=t dx \\ \frac{1}{t} du =dx \\ \int\limits_1^a \int\limits_0^\infty \frac{1}{t} f'(u) du dt \\ \int\limits_1^a \frac{1}{t } \int\limits_0^\infty f'(u) du dt \\ \int\limits_1^a \frac{1}{t} f(u)|_0^{z \rightarrow \infty} dt \\ \]
what is f(u)...
\[tx=u \\ \text{ so I think } f'(u)=\frac{\arctan(u)}{\frac{u}{t}}= t \frac{\arctan(u)}{u} \\ \text{ so I need \to evaluate } \\ \int\limits \frac{\arctan(u)}{u} du \\ y=\arctan(u) \\ \tan(y)=u \\ \sec^2(y) dy=du \\ \int\limits \frac{y}{\tan(y)} \sec^2(y) dy \\ \int\limits y \frac{\cos(y)}{\sin(y)} \frac{1}{\cos^2(y)} dy \\ \int\limits y \frac{1}{\sin(y) \cos(y)} dy \\ \int\limits 2y \csc(2y) d y \\ v=2y \\ dv =2 dy \\ \frac{1}{2} dv =dy \\ \frac{1}{2} \int\limits v \csc(v) dv\]
maybe I can't integrate that

- Kainui

Another favorite: \[\large \int\limits_0^\infty x^n e^{-x}dx\]

- xapproachesinfinity

i did IBP for the last one
\[I_k=n(n-1)(n-2).....(n-k)\int\limits_0^\infty x^{n-k-1}e^{-x}dx \]

- xapproachesinfinity

may be on the wrong truck

- Kainui

I woke up really early so I'm sorta tired. I'll let ya think about these and play around, I'll keep coming back tomorrow and give hints and more integrals of course if you like =P

- Kainui

I think you're on the right path for that last one @xapproachesinfinity and I think that will work, however there's another tricky way to do it too ;)

- xapproachesinfinity

hmm there is an easier way

- xapproachesinfinity

how about we look I1, I2..... and generalize
but that is gonna be by parts as well

- xapproachesinfinity

anyways have a good night rest :)

- xapproachesinfinity

it is definitely n! that last one

- xapproachesinfinity

doing it with cases give you the idea of the pattern that is raising

- xapproachesinfinity

somehow for the second i got \[I=\frac{\pi}{2}(1-a)\] for a>0

- anonymous

A recent favorite approach of mine: There a lot of integrals you can evaluate using the Laplace transform. Here's a good "model":
\[\int_0^\infty \frac{\sin^3 x}{x^2+1}\,dx\]
for natural \(k\). It involves introducing a parameter, as you would when differentiating under the integral sign, but there's no differentiation involved.
The hardest part is probably just reducing the numerator into "easier" trig functions via identities. The tedious part is taking the forward/inverse transforms to arrive at the result.

- xapproachesinfinity

for the second i send x to other side to to get I'=xI
not sure of this is valid lol

- xapproachesinfinity

then i manipulate the integral with arctan ax and artan x only

- xapproachesinfinity

does that sound doable?

- freckles

what did you do for second one?

- freckles

i'm still lost on the second one

- xapproachesinfinity

\[I=\int \frac{\arctan ax-\arctan x}{x}dx\]
\[xI=\int \arctan ax-\arctan xdx\]
try to do that integrand and then divide by x later

- xapproachesinfinity

oh no i just realized a mistake

- xapproachesinfinity

that was ridulucous of me lol

- xapproachesinfinity

perhaps Taylor would work

- xapproachesinfinity

gotta go :)
take care

- freckles

\[\int\limits f dx=F+C \\ \text{ where } F'=f \\ \text{ we have } \\ \int\limits_0^\infty \frac{\arctan(ax)-\arctan(x)}{x} dx \\ f(tx)=\arctan(tx) \\ \int\limits_0^\infty \frac{1}{x}[\arctan(tx)]_{t=1}^{t=a} dx \\ \\ \text{ Let } f(tx)=\arctan(tx) \\ \int\limits_0^\infty \frac{1}{x}[f(tx)]_{t=1}^{t=a} dx \\ \int\limits_0^\infty [\frac{f(tx)}{x}]_{t=1}^{t=a} dx \\ g(t)=\frac{f(tx)}{x} \\ \frac{dg}{dt}=\frac{1}{x} \cdot x \cdot \frac{df(tx)}{dt}=\frac{df(tx)}{dt} \\ \\ \]
\[\int\limits_0^\infty \int\limits_1^a \frac{dg}{dt} dt dx \\ \int\limits_1^a \int\limits_0^\infty \frac{dg}{dt} dx dt \\ \]
\[\int\limits_1^a [\frac{1}{t} f(tx)]_{x=0}^{x=z \rightarrow \infty } dt \\ \int\limits_1^a[\frac{1}{t}[f(t \cdot z)-f(t \cdot 0)] dt \\ \int\limits_1^a \frac{1}{t}[\arctan(tz)-\arctan(0)] dt \\ \int\limits_1^a \frac{1}{t}[\frac{\pi}{2}-0] dt \\ \ln(t)|_1^a \cdot \frac{\pi}{2} \\ [\ln(a)-\ln(1)] \frac{\pi}{2} \\ \frac{\ln(a) \pi}{2}\]

- freckles

and of course you already said a>0

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