Question

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.

Answer 1

Could be indefinite or definite! Whateves!

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

or maybe I just need to play with that idea more

Answer 6

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

Answer 7

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

Answer 8

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:

\[\large \int\limits_{-\infty}^\infty e^{-x^2}dx = \sqrt{\int\limits_{-\infty}^\infty e^{-x^2}dx \int\limits_{-\infty}^\infty e^{-y^2}dy} \]

Answer 9

\[I=\int\limits_0^\infty e^{-x^2} dx \\ I \cdot I=\int\limits_0^\infty e^{-x^2} dx \cdot \int\limits_0^\infty e^{-x^2} dx \\ \text{ change up the variable on one } \\ I^2=\int\limits_0^\infty e^{-x^2} dx \cdot \int\limits_0^\infty e^{-y^2} dy \\ I^2=\int\limits_0^\infty \int\limits_0^\infty e^{-x^2} e^{-y^2} dx dy\]

I don't know why the last two lines are equivalent?

Answer 10

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.

Answer 11

yes that is actually

Answer 12

\[I^2 =\int\limits_0^\infty \int\limits_0^\infty e^{-(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits_0^\infty \int\limits_0^\infty r e^{-r^2} dx dy \\ 0<x<\infty \\ 0<y<\infty \]
That is everything in first quadrant
so that means
\[0<r<\infty \\ 0<\theta<\frac{\pi}{2}\]
so we have
\[I^2=\int\limits_0^\frac{\pi}{2} \int\limits_0^\infty r e^{-r^2} dr d \theta \\ I^2=\int\limits_0^\frac{\pi}{2}[\frac{-1}{2}e^{-r^2}]_0^{ z \rightarrow \infty}] d \theta \\ I^2=\int\limits_0^\frac{\pi}{2}[\frac{-1}{2}e^{-z^2}+\frac{1}{2}] d \theta \\ I^2 = \int\limits_0^\frac{\pi}{2}\frac{1}{2} d \theta \\ I^2=\frac{1}{2} \frac{\pi}{2} \\ I=\frac{\sqrt{\pi}}{2}\]

So \[\int\limits_{- \infty}^\infty e^{-x^2} dx=2 \cdot I=2 \cdot \frac{\sqrt{\pi}}{2}=\sqrt{\pi}\]

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

\[I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{-(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{-r^2} dx dy \\ 0<x<\infty \\ 0<y<\infty \\ I^2=\int\limits\limits_0^\frac{\pi}{2} \int\limits\limits_0^\infty r e^{-r^2} dr d \theta \\ I^2=\int\limits\limits_0^\frac{\pi}{2}[\frac{-1}{2}e^{-r^2}]_0^{ z \rightarrow \infty}] d \theta \\ I^2=\int\limits\limits_0^\frac{\pi}{2}[\frac{-1}{2}e^{-z^2}+\frac{1}{2}] d \theta \\ I^2 = \int\limits\limits_0^\frac{\pi}{2}\frac{1}{2} d \theta \\ I^2=\frac{1}{2} \frac{\pi}{2} \\ I=\frac{\sqrt{\pi}}{2} \]

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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?

Answer 23

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

Answer 24

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: \[\large \int\limits_0^\infty \lfloor x \rfloor e^{-x} dx\]

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

@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?

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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!

Answer 33

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

Answer 34

@sithsandgiggles has asked a lot of integral questions

Answer 35

hard ones too

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

polar-ness*

Answer 41

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

Answer 42

i agree

Answer 43

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

Answer 44

\[\int\limits f'(tx) dt \\ \text{ Let } u(t)=tx \\ \frac{d(u(t))}{dt}=\frac{d(tx)}{dt} \\ \frac{d(u(t))}{dt}=x \\ \frac{du}{dt}=x \\ du=x dt \\ \frac{1}{x} du =dt \\ \int\limits f'(tx) dx=\frac{1}{x} \int\limits f'(u) du=\frac{1}{x}( f(u)+C)\]

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

Answer 45

I guess we can try the fubini theorem thingy

Answer 46

\[\int\limits_1^a \int\limits_0^\infty f'(tx) dx dt \\ \text{ Let } u(x)=tx \\ u'(x)=t \\ du=t dx \\ \frac{1}{t} du =dx \\ \int\limits_1^a \int\limits_0^\infty \frac{1}{t} f'(u) du dt \\ \int\limits_1^a \frac{1}{t } \int\limits_0^\infty f'(u) du dt \\ \int\limits_1^a \frac{1}{t} f(u)|_0^{z \rightarrow \infty} dt \\ \]
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

Answer 47

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

Answer 48

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

Answer 49

may be on the wrong truck

Answer 50

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

Answer 51

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 ;)

Answer 52

hmm there is an easier way

Answer 53

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

Answer 54

anyways have a good night rest :)

Answer 55

it is definitely n! that last one

Answer 56

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

Answer 57

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

Answer 58

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.

Answer 59

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

Answer 60

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

Answer 61

does that sound doable?

Answer 62

what did you do for second one?

Answer 63

i'm still lost on the second one

Answer 64

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

Answer 65

oh no i just realized a mistake

Answer 66

that was ridulucous of me lol

Answer 67

perhaps Taylor would work

Answer 68

gotta go :)
take care

Answer 69

\[\int\limits f dx=F+C \\ \text{ where } F'=f \\ \text{ we have } \\ \int\limits_0^\infty \frac{\arctan(ax)-\arctan(x)}{x} dx \\ f(tx)=\arctan(tx) \\ \int\limits_0^\infty \frac{1}{x}[\arctan(tx)]_{t=1}^{t=a} dx \\ \\ \text{ Let } f(tx)=\arctan(tx) \\ \int\limits_0^\infty \frac{1}{x}[f(tx)]_{t=1}^{t=a} dx \\ \int\limits_0^\infty [\frac{f(tx)}{x}]_{t=1}^{t=a} dx \\ g(t)=\frac{f(tx)}{x} \\ \frac{dg}{dt}=\frac{1}{x} \cdot x \cdot \frac{df(tx)}{dt}=\frac{df(tx)}{dt} \\ \\ \]


\[\int\limits_0^\infty \int\limits_1^a \frac{dg}{dt} dt dx \\ \int\limits_1^a \int\limits_0^\infty \frac{dg}{dt} dx dt \\ \]
\[\int\limits_1^a [\frac{1}{t} f(tx)]_{x=0}^{x=z \rightarrow \infty } dt \\ \int\limits_1^a[\frac{1}{t}[f(t \cdot z)-f(t \cdot 0)] dt \\ \int\limits_1^a \frac{1}{t}[\arctan(tz)-\arctan(0)] dt \\ \int\limits_1^a \frac{1}{t}[\frac{\pi}{2}-0] dt \\ \ln(t)|_1^a \cdot \frac{\pi}{2} \\ [\ln(a)-\ln(1)] \frac{\pi}{2} \\ \frac{\ln(a) \pi}{2}\]

Answer 70

and of course you already said a>0