How Good Am I at Integrals?

[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,

$$\int_{0}^{\infty}x^ne^{-x^2}dx =\frac{1}{2}\int_{0}^{\infty}z^{\frac{n+1}{2}-1}e^{-z}dz$$
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?

 

[Hootenanny]

Not technically difficult, but a nice introduction to recurrence methods;

$$I = \int{e^x\cdot\sin(x)}\; dx$$

 

[morphism]

$$\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \; dx$$

and

$$\int_0^{\pi / 2} \frac{1}{1 + \tan^{\pi} x} \; dx$$

are interesting.

 

[Eighty]

morphism: Neat, integrals I could do but not Mathematica. :)

 

[Gib Z]

Hootenanny - $$I = e^x \sin x - \int e^x \cos x dx$$
$$\int e^x \cos x dx = e^x \cos x + I$$
$$I= \frac{e^x(\sin x - \cos x)}{2}$$

morphism - Using the standard result $$\int^a_0 f(x) dx = \int^a_0 f(a-x)$$, your first integral is shown below:

$$I= \int^{\pi}_0 \frac{x\sin x}{1+\cos^2 x}dx = \int^{\pi}_0 \frac{\pi \sin x}{1+\cos^2 x}dx - I$$

$$2I = \pi \int^{\pi}_0 \frac{\sin x}{1+\cos^2 x} dx$$

Can't be bothered typing up all the latex, but basically u=cos x, and arctan form gets us :
$$I = \frac{\pi^2}{4}$$.

As for your second one, I find your torture methods interesting. We have a vacancy at Guantanamo Bay if you would like to apply.

 

[morphism]

Hehe

(Your solution for the first one is correct of course. Good job!)

 

[murshid_islam]

how do i solve morphism's second one?
$$\int_0^{\pi / 2} \frac{1}{1 + \tan^{\pi} x} \; dx$$

 

[morphism]

Use the "standard result" Gib posted (i.e. essentially the substitution u = pi/2 - x). Then mess around with things.

 

[Hootenanny]

Okay, one more trig one. I haven't looked through your previous post so I apologise if this has already been posted;

$$I = \int{\frac{dx}{4+\sin(x)}}$$

 

[Gib Z]

Nvm Much easier than I thought. I didn't even give it a try because I seemed to think the exponent of pi was a bad sign.
Let I be the original Integral.
$$I= \int^{\pi/2}_0 \frac{1}{1+\tan^{\pi} x} dx = \int^{\pi/2}_0 \frac{1}{1+\cot^{\pi} x} dx$$.

$$2I = \int^{\pi/2}_0 \frac{1}{1+\tan^{\pi} x} dx + \int^{\pi/2}_0 \frac{1}{1+\cot^{\pi} x} dx = \int^{\pi/2}_0 \frac{1}{1+\cot^{\pi} x} + \frac{1}{1+\tan^{\pi} x} dx$$. Simple cross multiplying and expansion gets us $$2I=\int^{\pi/2}_0 1 dx = \pi/2$$

Therefore $$I = \frac{\pi}{4}$$.

Nice ones morphism, got any more?

 

[Gib Z]

Umm Hootenannys one took a while, I Basically did t=tan (x/2) and made it a rational function so on so forth.

I got $$\frac{2}{\sqrt{15}} \arctan \frac{4 \tan (x/2) +1}{\sqrt{15}} + C$$ which seems like a hairy answer, I don't think I am right. I can't be stuffed to differentiate it to check lol.

 

[morphism]

Nice ones morphism, got any more?

Here you go:

$\int_0^{\pi / 2} \tan^{-1} x + \cot^{-1} x \; dx$

These are inverses not reciprocals. (N.B.: There exists a solution that is one line long.)

$\int_0^1 \frac{\log(x + 1)}{x^2 + 1} \; dx$

$\int_0^1 \frac{\log(x + 1)}{x} \; dx$

$\int_0^\pi \log (\sin x) \; dx$

 

[Gib Z]

Hehe the first one is easier than I thought. I first tried u = arctan x + arccot x, but when I tried to find du I realized the derivative was 0, so I finally noticed it was a constant function. I tried x=1, the constant function is equal to pi/2, so the solution is $$\frac{\pi^2}{4}$$!

I couldn't do the 2nd one :(

Third one I did u= log (2), du/dx=1/x which is in there so all good.
So I integrated it and ended up with $$\frac{\log^2 (x+1)}{2}$$.

Last one, I split it up into 2 integrals, from 0 to pi/2, and pi/2 to pi.
Using symmetry arguments and the such, I got that it $$2I= \int^{\pi/2}_0 \log(\sin x) dx = \int^{\pi/2}_0 \log(\cos x) dx$$ but I am stuck from there.

 

[TD]

Use the property of logs that log(xy) = log(x) + log(y) with the double angle formula of the sine.

 

[Gib Z]

$$I=\frac{1}{4} \int^{\pi/2}_0 \log_e (\frac{1}{2}\sin 2x) dx$$...I don't get it...

 

[Gib Z]

Wait up I think I am getting it, Ill post it in 5 mins

 

[Gib Z]

Nope lost again...

 

[Hootenanny]

Umm Hootenannys one took a while, I Basically did t=tan (x/2) and made it a rational function so on so forth.

I got $$\frac{2}{\sqrt{15}} \arctan \frac{4 \tan (x/2) +1}{\sqrt{15}} + C$$ which seems like a hairy answer, I don't think I am right. I can't be stuffed to differentiate it to check lol.

Absolutely spot on Gib, it can't be simplified any further

 

[morphism]

Hehe the first one is easier than I thought. I first tried u = arctan x + arccot x, but when I tried to find du I realized the derivative was 0, so I finally noticed it was a constant function. I tried x=1, the constant function is equal to pi/2, so the solution is $$\frac{\pi^2}{4}$$!

Yes, that's right, but there's a solution that'st truly one ling long. Yours involves differentiating things and evaluating it at a point, so it's longer than one line, especially if you don't know the derivatives of those things. The solution I have in mind is: Notice that arctan(x) and arccot(x) over [0, pi/2] correspond to the two accute angles in a right-angled triangle. So they add up to pi/2.

I couldn't do the 2nd one :(

How about setting x=tan(t)?

Third one I did u= log (2), du/dx=1/x which is in there so all good.
So I integrated it and ended up with $$\frac{\log^2 (x+1)}{2}$$.

I'm not sure I follow. Do you mean you set u=log(x)? That doesn't help...

Last one, I split it up into 2 integrals, from 0 to pi/2, and pi/2 to pi.
Using symmetry arguments and the such, I got that it $$2I= \int^{\pi/2}_0 \log(\sin x) dx = \int^{\pi/2}_0 \log(\cos x) dx$$ but I am stuck from there.

That's good. Now use these to get I. (sin(x) = 2sin(x/2)cos(x/2) and log(xy) = log(x)+log(y).)

 

[Gib Z]

The third one I sed u = log(x+1), and that helps because du/dx = 1/x, which is there as well. So My solution for it was $$\frac{\ln 2)^2}{2}$$.

Last one, that's what i did and i got this!

$$I=\frac{1}{4} \int^{\pi/2}_0 \log_e (\frac{1}{2}\sin 2x) dx$$

Can't do that one!

 

[Gib Z]

For the one you said x=tan t, I did it but I can finish off $$\int \ln (\tan t +1) dt$$

 

[morphism]

The third one I sed u = log(x+1), and that helps because du/dx = 1/x, which is there as well. So My solution for it was $$\frac{\ln 2)^2}{2}$$.

But du/dx isn't 1/x! It's 1/(x+1).

Last one, that's what i did and i got this!

$$I=\frac{1}{4} \int^{\pi/2}_0 \log_e (\frac{1}{2}\sin 2x) dx$$

Can't do that one!

OK, let's see.

$$I = \int_0^{\pi} \log(2) + \log(\sin \frac{x}{2}) + \log(\cos \frac{x}{2} ) \; dx$$

From your observations, we can see that this is just I = pi log(2) + I + I.

 

[Gib Z]

MY GOD KILL ME, I really should have seen those >.< Staring me in the face..

Okie well its getting late here in Australia, Ill try again with the remaining 2 tomorrow.

 

[morphism]

For the one you said x=tan t, I did it but I can finish off $$\int \ln (\tan t +1) dt$$

$$\ln (1 + \tan t) = \ln(\sin t + \cos t) - \ln(\cos t) = \ln \left( \sqrt2\cos(t - \frac{\pi}{4}) \right) - \ln(\cos t)$$

Good night!

 

[Gib Z]

Ok I am back But for your hint on the x=tan t one, am I just awfully stupid or is that still quite hard to integrate..

 

[morphism]

Here's the next step:

$$\ln(\sqrt 2) + \ln \left( \cos \left( t - \frac{\pi}{4} \right) \right) - \ln(\cos t)$$

Now use the limits of integration...