freckles
  • 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.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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freckles
  • freckles
Could be indefinite or definite! Whateves!
Kainui
  • 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
  • freckles
I think that one is a famous one. I think I remember seeing it done by taylor series.

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freckles
  • 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
  • freckles
or maybe I just need to play with that idea more
freckles
  • freckles
nevermind don't think it possible with taylor or power whatever you call it
Kainui
  • Kainui
Haha yeah there's a nice way of doing it. =)
Kainui
  • 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
  • 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
  • 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
  • freckles
yes that is actually
freckles
  • 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
  • Kainui
Awesome, alright I guess I need to find something harder to send your way.
freckles
  • freckles
lol when you say harder I know it will be because you are crazy good at math
freckles
  • freckles
by the way I did make a mistake in one of my lines
freckles
  • 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
  • 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
  • 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
  • freckles
yeah that is the same I got I'm just saying I put the r in too early
Kainui
  • 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
  • freckles
now this one I do believe is entirely new to me let me think a bit
freckles
  • 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
  • freckles
you know what nevermind it is probably some kinda pattern or whatever
Kainui
  • 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
  • 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
  • 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
  • xapproachesinfinity
integral party lol i have seen that first integral many times ant its value never question how it is root(pi)
freckles
  • 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
  • freckles
and wow you guys did put a lot of work into that one problem
freckles
  • freckles
anonymous
  • 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
  • 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
  • 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
  • freckles
@sithsandgiggles has asked a lot of integral questions
freckles
  • freckles
hard ones too
freckles
  • 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
  • freckles
though I don't think that is exactly what I want
xapproachesinfinity
  • 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
  • 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
  • freckles
polar-ness*
freckles
  • freckles
you know like since we know x^2+y^2=r^2
xapproachesinfinity
  • xapproachesinfinity
i agree
freckles
  • freckles
then eventually using the r dr dtheta= dx dy thing so the integration actually becomes doable
freckles
  • 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
  • freckles
I guess we can try the fubini theorem thingy
freckles
  • 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
  • Kainui
Another favorite: \[\large \int\limits_0^\infty x^n e^{-x}dx\]
xapproachesinfinity
  • 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
  • xapproachesinfinity
may be on the wrong truck
Kainui
  • 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
  • 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
  • xapproachesinfinity
hmm there is an easier way
xapproachesinfinity
  • xapproachesinfinity
how about we look I1, I2..... and generalize but that is gonna be by parts as well
xapproachesinfinity
  • xapproachesinfinity
anyways have a good night rest :)
xapproachesinfinity
  • xapproachesinfinity
it is definitely n! that last one
xapproachesinfinity
  • xapproachesinfinity
doing it with cases give you the idea of the pattern that is raising
xapproachesinfinity
  • xapproachesinfinity
somehow for the second i got \[I=\frac{\pi}{2}(1-a)\] for a>0
anonymous
  • 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
  • xapproachesinfinity
for the second i send x to other side to to get I'=xI not sure of this is valid lol
xapproachesinfinity
  • xapproachesinfinity
then i manipulate the integral with arctan ax and artan x only
xapproachesinfinity
  • xapproachesinfinity
does that sound doable?
freckles
  • freckles
what did you do for second one?
freckles
  • freckles
i'm still lost on the second one
xapproachesinfinity
  • 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
  • xapproachesinfinity
oh no i just realized a mistake
xapproachesinfinity
  • xapproachesinfinity
that was ridulucous of me lol
xapproachesinfinity
  • xapproachesinfinity
perhaps Taylor would work
xapproachesinfinity
  • xapproachesinfinity
gotta go :) take care
freckles
  • 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
  • freckles
and of course you already said a>0

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