Improper Integrals: How to solve these problems using Comparison Thm or if possible Evaluation?
817021
I will try to show you that the first integral is non convergent:\nf(X)=e^(x-e^(-x))\nFirst we realize the function can never be negative for any x є R because e^x will never be negative for all x є R : x < 0 => e^x >0 , x=0 => e^x =1 x> 0 => e^x >0\nTherefore the integral will always be positive if its taken from b to a where b < a .\nLet try to rewrite the expression:\ne^(x-e^(-x)) = e^x * e^(-e^(-x))\nLets look at each term at the time:\nFirst\n1) e^x\na) . . x → + ∞ => e^x → + ∞\nb). . .x→ - ∞ => e^x → 0 \n \nThen\n\n2) e^(-e^(-x))\nfirst Look at the exponent to e i.e "-e^-x = -1/e^x "\na). . .x →+ ∞ => -1/e^x → 0(-) from the left\nb). . .x → - ∞ => -1/e^x → - ∞\n\nNow look at the hole expression 2) "e^(-e^(-x))"\naa). . .x → + ∞ => e^(-e^(-x)) → e^0(-) = 1\nbb). . x → - ∞ => e^(-e^(-x)) → (e^-∞) =0\n\nNow we can combine the results.\nFrom 1)a &2)aa we have: \nx → + ∞ => e^x * e^(-e^(-x)) → (+ ∞ )* 1 → ∞\nfrom 1)b & 2)bb we have:\nx → - ∞ => e^x * e^(-e^(-x)) → 0 * 0→ 0\nHence: the function is always positive with no upper limit for x going to + infinity there fore the integral cannot be convergent.\n======================================\nThe next function you can treat the same way, but the function ln(x) is only defined for x>0 so we have the domain x Є (0, +∞] ( I hope this is the way you are used to write it)\n\nf(x) = (x^1/2)*(ln(x))/(x+1)(x^2+x+1) we separate the equation in two parts\n\n1) \nf(x)=(x^1/2)/(x+1) → x^½  for x → 0(+) (from right) because denominator will never be < 1\nf(x)=(x^1/2)/(x+1) → 1/(x+1) for x → +∞ because the denominator is a must stronger function.\n\n2)\nln(x)/ (x^2+x+1) → ln(x) → - ∞  for x → 0(+) again the denominator will never be <1  \nln(x)/ (x^2+x+1) →1/ (x^2+x+1) → 0(+)  for x → + ∞ because ln(x) is must weaker then any polynomials with n >1 for x going to infinity \n\nCombining 1) and 2)\nf(x) → x^½  * ln(x) → - ∞ for x going to 0(+)\nf(x) →→ 1/[(x+1) (x^2+x+1)] → 0 for x →∞ +\n\nAbout the last function, I am not, able to say whether the integral is convergent or divergent. \nsorry!
748051
The primitive of your integral is:\n\ns*(e^(st))*(1/w)*atan(s/w)\n\nYou have to do a trigonometric substitution to get there. Do not always trust the weird answers you get with complex calculators.
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
480262
the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated
352249
This integral does exist (it can be evaluated for any endpoints you choose). Since it cannot be represented as a finite combination of elementary functions and operations, it is given its own name, the error function. Many integrals like this have their own names and are quite useful in applied mathematics. See http://mathworld.wolfram.com/Erf.html .\nTo say that it does not exist is unrigorous.
168841
Suppose that the limits on your integration are from 1 to X\n\nint{1/x dx from 1 to X} = (1-1)+int{1/x dx  from 1 to X}\n\nand one finds out that 0=0\n\nIf you want to perform an indefinite integral,\n\nint{udv}=uv-int(vdu} +C, where C is a constant.\n\nif u=1/x, and v=x, then choosing C=-1 makes this equality true.
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
71557
integral of [x^y+y^x]dx= [x^(y+1)]/(y+1)+(y^x)*(1/ln y)+constant(if not definite)
815728
first this is not a PDE but ODE\n\nsimple integration would do
421543
use substitution:\n\nlet u=-x^2, then du= -2x dx, so 4x dx= -2du \n\nso now we have the integral of -2e^u wrt to u\nwhich is:\n\n-2e^u\n\nnow putting back in that u=-x^2\n\nwe have the answer:\n\n-2e^(-x^2)
274471
i made a little comment for you on your integral answer:)
832512
either square root or erauqs
772619
integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
74089
http://integrals.wolfram.com/
501273
Didn't got the question do you want the solution for all the problems of goodman's Introduction to Fourier optics book? Well I do not think I have them any more if you want a specific problem I can help you. I can even try to get a chapter solution for you. Just tell me each one, I can try to get it for you, I do not garantee but I think I may have a way to got them just send me an email (I have a yahoo email). If you have the solutions and you don't understand tell me each problem I can do it in more detail probably, not garantee at 100% but at a good 90% .
83265
That's an elliptic integral.  Check out the link.
888834
Integrate from -π/2 to π/2 and double it