Why  $\int e^{-x^2} dx$ cannot be expressed in a closed form? Definite integration of this function is possible however I could not understand why indefinite Integration is not possible.
120843
I have seen this claim on Wikipedia and other places. Which branch of mathematics does this result come from?
95025
Find the indefinite integral  $$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$  Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..
32378
I'm trying to solve the following problem     $$\text{Evaluate:} \lim_{b \to 1^+} \int_1^b \frac{dx}{\sqrt{x(x-1)(b-x)}}$$   I'm tempted to think it's 0 because the bounds would be about equal, but that's not the correct answer. But I don't know what integration techniques to use, so any hints are greatly appreciated.  I don't know how to solve for the indefinite integral itself.
116454
Given that $\int_{-\infty}^\infty e^{-x^2} \, dx=\sqrt{\pi}$. Find the value of $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+xy+y^2)} \, dx\,dy$$  I don't understand how I find this double integral by using the given data. Please help.
95181
In multivariable calculus, definite integrals with multiple variables seem routine. However, I have not seen any example of an indefinite multiple integral. In fact, it seems as if limits of integration are a necessary part of multiple integrals. Is there actually any way to evaluate indefinite integrals of functions with more than one variable?
107759
Compute $$ \int_{0}^{ \infty }\frac{\ln xdx}{x^{2}+ax+b}$$ I tried to compute the indefinite integral by factorisation and partial fraction decomposition but it became nasty pretty soon. There must be another way to directly evaluate it without actually computing the indefinite integral which I don't know!
57753
I was asked to solve this indefinite integral using Integration by parts.  $$\int \sqrt{1-x^2} dx$$  I know how to solve if use the substitution $x=\sin(t)$ but I'm looking for the Integration by parts way.   any help would be very appreciated.
74727
These integrals are supposed to have an elementary closed form, but Mathematica only returns something in terms of elliptic integrals. I got them from the book Treatise on Integral Calculus by Edwards. How can we evaluate them?     $$ I = \int{\frac{\sqrt{1+x^4}}{1-x^4}dx}\\ J = \int{\frac{x^2}{(1-x^4)\sqrt{1+x^4}}} dx  $$
11177
$ \int_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}} dx $ = -$  2.887270... $  The function has no antiderivate and there's no symmetry here to help, if you solve this I would be thankful if you would for a highschooler ^^
33079
This integral has been bugging me since yesterday:     $$\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$$   I've tried substitution $y={1}/{x}$ and $e^y={1}/{x}$, but those didn't help much. Wolfram Alpha gives me result: $-0.67172$. Could anyone here please help me to obtain the closed form of the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
122196
How $$\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$$ exists?   It is difficult question to me. i have tried to evaluate by using fact that $$\int_{-\infty}^{\infty} f(x) \ dx =\int_{-\infty}^{0} f(x)\, dx + \int_{0}^{\infty} f(x)\, dx$$ but i have failed in this one.   any hints on this one?
66971
I am told that an expression like $$ \int_a^x f(x)dx $$ is not well formed, i.e. it should be $$ \int_a^xf(t)dt $$ or similar.  Why is it that the limits of integration can't depend on the variable of integration?
107304
The problem is to show  $$\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$$  for $t \gt 0$.  I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution $u=e^{-x}$ and came to a new integral (hopefully after no mistakes)  $$ \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, $$  but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself.  Thanks a lot!
73548
Can we find a closed form for this integral: $$\int_0^\infty e^{-ix} x^{\alpha -1} \mathrm{d}x$$  I am assuming it requires complex analysis and the use of Cauchy's residue theorem, but is there any elementary way other than complex analysis to evaluate this? I am not asking for a solution; a hint or guidance would not be overlooked. Thank you in advance!
53688
I'm trying to solve this integral: $$\int_{-\infty}^\infty e^{-x^2-x}dx$$ WolframAlpha shows this to be approximately $2.27588$. I tried to solve this by integration by parts, but I just couldn't get there. I'd be glad if someone could show me how to do it.   I've included my attempt at the problem below:  Note $\int_{-\infty}^\infty e^{-x^2-x} = \int_{-\infty}^\infty e^{-x^2}e^{-x}dx$. Then we can integrate by parts. Let $u(x) = e^{-x^2}$. Then $u'(x) = -2xe^{-x}$. Let $v'(x) = e^{-x}$. Then $v(x) = -e^{-x}$.  Then $u(x)v'(x) = u(x)v(x) - \int v(x)u'(x)dx$, i.e. $$\int_{-\infty}^\infty e^{-x^2}e^{-x}dx = -e^{-x^2}e^{-x} - \int e^{-x}2xe^{-x}dx = -e^{-x^2-x} - 2\int e^{-2x}xdx$$  Then we integrate $\int e^{-2x}xdx$ by parts. We pick $u(x) = x$, $u'(x) = dx$, $v'(x) = e^{-2x}$, and $v(x) = \int e^{-2x} dx = \frac {-1}{2} e^{-2x}$ by u-substitution. Skipping some steps, it follows that  $$\int e^{-2x}xdx = -\frac{1}{4}(e^{-2x})(2x+1)$$ Then  $$ -e^{-x^2-x} - 2\int e^{-2x}xdx = -e^{-x^2-x} + \frac{1}{2}(e^{-2x})(2x+1)$$ Which, when evaluated numerically, doesn't yield the desired result. So there's a problem here. Maybe someone else knows how to do this.
17907
Most of us are aware of the classic Gaussian Integral   $$\int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$  I would be interested in evaluating the similar sum  $$\sum_{x=0}^\infty e^{-x^2}$$  Now, because $\exp(-\lfloor x \rfloor^2) \ge \exp(-x)$, we find   $$\sum_{x=0}^\infty e^{-x^2}= \int_0^\infty e^{-\lfloor x \rfloor^2}\, dx \ge \int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$  Does a closed form for this sum exist?  If so, what would it be?  I would be very interested in how a closed form would be found for this function.
29881
Please consider the integral:  $$\int_{-a}^{a}x^2dx=\frac{2a^3}{3}$$  I would like to know why I can't make the substitution:  $$u=x^2$$  When I make the substitution, the limits of the integral will be the same, and the integral itself will be zero, which is the wrong answer. So why does this simple change of variables not work as I have expected?  Please note that I do not want help solving the integral, I know how to solve it several ways. My question is why does this specific attempt at a solution not work?