Are You Ready to Challenge Your Integral Solving Skills?

[ssd]

how did you get this:

This is a standard result
$$\int_{0}^{a}f(x)dx= \int_{0}^{a}f(a-x)dx$$

 

[Gib Z]

If you say so?...I've never seen that before...

 

[murshid_islam]

ssd, can you show me how to prove this?:

$$\int_{0}^{a}f(x)dx= \int_{0}^{a}f(a-x)dx$$

edit: ok i think i got it. by substituting a - x = u, i get,
$$\int_{0}^{a}f(a-x)dx = -\int_{a}^{0}f(u)du = \int_{0}^{a}f(x)dx$$

 

[Gib Z]

And that proof works how? I can't see how the 2nd part equals the 3rd.

 

[cristo]

And that proof works how? I can't see how the 2nd part equals the 3rd.

The variables u and x are just (dummy) integration variables, so from going from the second part to the third, let u=x. Then noting that $$\int_a^bf(z)=F(b)-F(a)=-[F(a)-F(b)]=-\int_b^a f(z)$$ where F(z) is an antiderivative of f(z), yields the result.

 

[murshid_islam]

$$\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta<br /> =\int_{0}^{\pi}\frac{\pi\sin\theta}{1 + \cos^4\theta}d\theta<br /> - \int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta$$
Move the 2nd integral to left to get 2I, I=the original integral.
For the first integral put z=$$cos\theta$$

even then it becomes messy. i get,
$$I = \pi \int_{0}^{1}\frac{dz}{1 + z^4}$$
i have read that even leibniz found this cumbersome. i have to complete the squares in the denominator, then split into partial fractions and so on...

 

[siddharth]

I don't think a trig sub will work. Try writing the denominator as $(u^2-\sqrt 2u+1)(u^2+\sqrt 2u+1)$, then use partial fractions. It should work, but it won't be pretty!

There's a pretty way to do that integral. Involves some trickery.

https://www.physicsforums.com/showthread.php?t=127563

 

[ssd]

even then it becomes messy. i get,
$$I = \pi \int_{0}^{1}\frac{dz}{1 + z^4}$$
i have read that even leibniz found this cumbersome. i have to complete the squares in the denominator, then split into partial fractions and so on...

The process is cumbersome no doubt but the problem is such...By the way I found it interesting.

 

[ssd]

And that proof works how? I can't see how the 2nd part equals the 3rd.

$$\int_{a}^{b}f(u)du = \int_{a}^{b}f(x)dx = F(b)-F(a)$$
where, f(y) is the first derivative of F(y) w.r.t. y.

 

[Gib Z]

K thanks guys, got it. As for that integral you have left, I feel sorry for you >.<

 

[murshid_islam]

The process is cumbersome no doubt but the problem is such...By the way I found it interesting.

i found the integral interesting too, even if it is cumbersome.

 

[Gib Z]

O well, let's revive this thread again. Anybody have another, single variable, indefinite integral? It doesn't even have to be from a textbook, it can be your homework, or one your having trouble with :D

 

[murshid_islam]

here is an extremely easy one:

$$\int_{-1}^{1}x^3e^{-x^{4}}\cos 2xdx$$

although it is very easy, i found it interesting

 

[Gib Z]

Ahh I am thinking integration by parts, $u=x^3e^{-x^4}$, and dv = cos 2x, and integration by parts again on the u, but could you just tell me if I am right before I do it, i don't want to spend so much time on it if it won't get me anywhere lol.

 

[gammamcc]

0

There are hard ways and easy ways. (Try graphing it. Notice any symmetry?)Try

\int_-90000 ^90000 sin^57(x) e^{-\pi^{77} 234535x^88}cos(2453245x^4) (x^{99999994}+1)dx

 

[d_leet]

Ahh I am thinking integration by parts, $u=x^3e^{-x^4}$, and dv = cos 2x, and integration by parts again on the u, but could you just tell me if I am right before I do it, i don't want to spend so much time on it if it won't get me anywhere lol.

It might be doable that way, but I doubt it. Note that this is a definite integral, what is special about the bounds? What is special about the function?

 

[Gib Z]

:D:D:D

Thanks a heap guys, I can't believe I didn't see that. I already noticed it was an odd function, but for some reason didn't make the connection.

The Answers 0 :).

As for gammamcc's one, I am just going to put the tex brackets that he forgot to over here, because I don't understand it written like that, then ill try it.

$$\int_{-90000}^{90000} \sin^{57} (x) e^{-\pi^{77} 234535x^{88}}\cos (2453245x^4) (x^{99999994}+1) dx$$

 

[Gib Z]

Thats an odd function as well...so the answers zero again...

 

[murshid_islam]

Ahh I am thinking integration by parts, $u=x^3e^{-x^4}$, and dv = cos 2x, and integration by parts again on the u, but could you just tell me if I am right before I do it, i don't want to spend so much time on it if it won't get me anywhere lol.

you don't need any calculations at all. that's why i said it is extremely easy.
just look at the function $$x^3e^{-x^{4}}\cos 2x$$. do you see anything? is it odd? even? then you can immediately see what the answer is.

edit: oh, sorry, you have already got help. didn't notice that.

 

[Gib Z]

And as I just realized, that integral is not expressible in terms of elementary functions...

EDIT: INDEFINITE integral i meant to say.

 

[murshid_islam]

And as I just realized, that integral is not expressible in terms of elementary functions...

EDIT: INDEFINITE integral i meant to say.

how did you realize that?

i have another integral. i don't know how to do it. can anybody help?

$$\int_{0}^{\infty}x^ne^{x^2}dx$$

 

[Gib Z]

I worked out the taylor series and integrated term by term, this is my best:

$$\int_0^{\infty} x^{a} e^{x^2} dx =\lim_{x\rightarrow {\infty}} \sum_{n=0}^{\infty} \frac{x^{a+(2n+1)}}{n!\cdot (a+(2n+1))}$$

That doesn't help much

 

[arildno]

how did you realize that?

i have another integral. i don't know how to do it. can anybody help?

$$\int_{0}^{\infty}x^ne^{x^2}dx$$

It diverges.

 

[Gib Z]

Looking at the summation again, It looks like it does lol.
EDIT: Or one could see that both are increasing functions, which i just noticed and am starting to feel stupid about for not noticing sooner.

 

[murshid_islam]

what about $$\int_{0}^{\infty}x^ne^{-x^2}dx$$

 

[ssd]

what about $$\int_{0}^{\infty}x^ne^{-x^2}dx$$

Putting z=x^2,

$$\int_{0}^{\infty}x^ne^{-x^2}dx<br /> =\frac{1}{2}\int_{0}^{\infty}z^{\frac{n+1}{2}-1}e^{-z}dz$$
Apart from the factor 1/2, the integral is a Gamma Integral with parameters (n+1)/2,1.
If (n+1)/2 is a positive integer, then the integral = 0.5[(n-1)/2]! ... (Gamma distribution is known as Erlang distribution if the shape parameter is an integer.)

 

[murshid_islam]

what if (n+1)/2 is not an integer?

 

[ZioX]

It becomes $$\frac{1}{2}\Gamma(\frac{n+1}{2})$$.

Which can be approximated...no closed form (in general).

 

[Gib Z]

Closed forms can be generated for any integer n with the following identities:

$$\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{1-2z} \; \sqrt{\pi} \; \Gamma(2z). \,\!$$

And

$$\Gamma\left(\frac{n}{2}+1\right)= \sqrt{\pi}\, \frac{n!}{2^{(n+1)/2}}$$ (n odd)

 

[Gib Z]

Lol its been a while since I've posted here. Anybody got an interesting integral?