In the WP article about , there is an integral solved as: $$K(x,x';t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}dk\,e^{ik(x-x')} e^{-\frac{i\hbar k^2 t}{2m}}=\left(\frac{m}{2\pi i\hbar t}\right)^{\frac{1}{2}}e^{-\frac{m(x-x')^2}{2i\hbar t}}$$ I was trying to find out how to evaluate the integral, using the method of &quot;completing the squares&quot; at the exponential. But at the end, I came up with $\gamma\int_{-\infty}^{\infty}e^{i\alpha (z-\beta)^2}dz$, where $\gamma$, $\alpha$ and $\beta$ are constants. The answer of the WP can be reached if it is solved as a gaussian integral: $= \gamma \sqrt{\frac{\pi}{i\alpha}}$ But is it valid to use the formula of the gaussian integral for a complex constant? When I look at the main integral (of K), the exponentials are after all oscillating functions, that don't have necessarily  to go to zero at infinity.
24570
Although this question is going to seem completely trivial to anyone with any exposure to path integrals, I'm looking to answer this precisely and haven't been able to find any materials after looking for about 40 minutes, which leads me to believe that it makes sense to ask the question here. In particular I'm looking for an answer wherin any mathematical claims are phrased as precisely as possible, with detailed proofs either provided or referenced. Also my search for a solution has led me to think I'm actually looking for a good explenation of Wick rotation, which I can't really claim to understand in detail. Any good references about this would be very welcome as well.  I'm looking to make sense of the following integral identity:  $$\int_{-\infty}^{\infty} dx \ \exp\left(i\frac{a}{2}x^2+iJx\right)=\left(\frac{2\pi i}{a}\right)^{1/2}\exp\left(\frac{-iJ^2}{2a}\right), \qquad a,J\in\mathbb{R}$$   (and various other sources) say that "This result is valid as an integration in the complex plane as long as a has a positive imaginary part." Clearly the left hand side does not exist in Lebesgue sense for real $a, J$. An answer to the question "" claims:      "it is convergent as a Riemann integral, thanks to some rather delicate cancellations. To make the integral well defined -- equivalently to see how these cancellations occur -- we need to supply some additional information.   Wick rotation provides a way of doing this. You observe that the left hand side is analytic in t   , and that the right hand side is well-defined if Im(t)&lt;0. Then you can define the integral for real t by saying that it's analytic continued from complex t with negative imaginary part."   I want to see the gory details and all known motivation for the validity of this procedure for the kinds of applications where such integrals occur. Suggestions such as "include an $i\epsilon$ in order to make it finite" seem arbitrary. In that case one would have to motivate that prescription from the very start, that is within the modeling procedure that ends up giving that integral expression (which is likely the correct way to approach this). I'm also not sure how to interpret the right hand side, since it involves the square root of an imaginary number, which should involve some choice of branch cut, which I have never seen specified in connection to this formula.
101742
I need solve this integral, and I tried various methods of solving and did not get it. The integral is:   $$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta,$$  where $t$ is a positive integer.
24552
I am solving the following integral: $$\int_0^\infty \frac{x}{1+x^3}dx$$ I need to solve it using integration and residue theorems. I tried to convert it to complex function, for example $\frac{z}{1+z^3}$, but it is not even, so it is difficult to find the curve for integration.   Any hints are welcome!  Thank you!
112459
It recently occurred to me that I did not know how to solve equations of the form $k^x=x^c$ for any two constants $k$ and $c$. After much pain in algebraically manipulating the equation (using logarithms) I confirmed the x is trapped in an exponent and cannot be extracted without some external knowledge.  Finding the roots of the derivative creates a similar problem.   I then did some research and discovered Lambert's W function, which I haven't the mathematical maturity to understand deeply. ().   I'm looking for some rigorous intuition towards why without numerical methods or Lambert's W function the equation is not solvable for x, or more specifically, a proof as to why this class of function is unsolvable using only logs and rationals? The best answer I have for myself is that because x is both a power and an exponent, either extracting x with a log or removing the exponent with a rational will keep the other side of the equation unsimplified. What is the mathematical language with which to express this idea?  As an aside, I'm also wondering what area of mathematics considers issues such as these (this one being on the simpler side).
42690
how to calculate $$\int_0^\infty \frac{x^{3}}{e^{x}-1} \; dx$$   Be $q:= e^{z}-1 , p:= z^{3}$ , then $e^{z} = 1 $ if $z= 2\pi n i $, so the residue at 0 is : $$\frac{p(z_{0})}{q&#39;(z_{0})} = 2\pi i n ^{3}$$  problem is that this is not symmetric, so how does one find the definite integral?
92455
$$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$  I am not sure how to solve this question. Can anyone help me to approach this problem. Thanks.
21986
I want to evaluate the integral   $$\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}e^{i t x} dx$$   for all $t \in \mathbb{R}$. I would preferably do it using the tools of complex analysis, but since I haven't found any other thread dealing with this integral at all so far, I'd be open to any approach.  I must admit that I haven't really found a way to start. The function $f(z) := \frac{\sin^2(z)}{z^2}e^{i t z}$ has an isolated singularity (a removable singularity, with continuation $f(0) = 1$, to be precice) at $z = 0$. I think a clever way would be to integrate over the edge of rectangle, semicircle or "slice of cake" (the edges of which we then would let go against $-∞$ and $∞$), then maybe somehow apply the Cauchy theorem or Residue theorem. But I don't know how to concretely do that, or how to even get started with that.  On a more general note, I'm having trouble understanding how to transform a "normal" integral over a real interval into one where we integrate over the edges of a geometric shape; and I also don't know, based on which pattern one can see with geometric shape fits best. Are there any tips or hints on how to get more intuition for that? (Apart from, obviously, dealing with these integrals regularly, what I'm trying to do but where I'm still failing a lot of times.)  I also noticed that the above integrals looks strikingly similar to the Fourier transformed of $f$ (or, even more similar to the formula for the inverse transformed). So maybe this could help us with this integral? I'm not really sure though.
94794
I want to solve the integral .Solve is difficult. I want to use statistical methods to solve them. $$\int_{0}^{+\infty}x \exp\{ ax-b x^2\}d x=\int _{0}^{+\infty} x\exp\{-b(x^2-\frac{a}{b}x)\}dx=\\ exp\{\frac{a^2}{4b^2}\}\int_{0}^{+\infty}x\exp\{-b(x-\frac{a}{b})^2\}dx=\sqrt{\frac{\pi}{4}}\exp{\frac{a^2}{4b^2}}\int_{0}^{+\infty}\frac{\sqrt{2}}{\pi(1\2)}x \exp\frac{-b}{2(1\2)}(x-\frac{a}{b})^2dx=\frac{\pi}{4}\exp\{\frac{a^2}{4b^2}\}.E(HN) $$ Wherever a and b are fixed and HN half normal distribution. What is the final answer
111338
Asked a question to evaluate the Gaussian Integral,  $$\dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty x^2 \exp(-x^2/2) dx  $$  using the the following approximation,  $J=\Bbb E[X^2] \sim J_N = 1/N \sum_1^NX_n^2 $  where $x_n \sim N(0, 1)$  Examin $J_N$ for different values of $N$ and plot an error graph (difference between exact and numerical values)  I can simulate and generate the approximations but I am stuck at getting the exact value to compare to. As far as I am aware there are no analytical solutions for gaussian integrals. The other way I can generate comparable numbers is to go use numerical integration techniques but those are approximate values too.  Any help will be greatly appreciated!
99328
What are some different methods to evaluate  $$ \int_{-\infty}^{\infty}  x^4 e^{-ax^2} dx$$  for $a &gt; 0$.  This integral arises in a number of contexts in Physics and was the original motivation for my asking. It also arises naturally in statistics as a higher moment of the normal distribution.  I have given a few methods of evaluation below. Anyone know of others?
111197
$$\int_{-\infty}^{\infty} \frac{\cos(mx) dx}{e^{-x}+e^x} = \frac{\pi}{e^{m\pi /2}+e^{-m\pi /2}}$$  I need to evaluate the above integral specifically using a rectangluar contour and at some point applying the residue theorem. I should be ok with the residue part, but I'm confused at to how to get this into the example form shown in the book:  $$I =\int_{-\infty}^{\infty} \frac{e^{ax}dx}{1+e^x}=\frac{\pi}{\sin(a\pi)} $$  I tried using the identity $\cos (mx) = \frac{e^{ix}+e^{-ix}}{2}$ but I end up with $\frac{1}{2} \int_{-\infty}^{\infty}\frac{e^{ix}(e^{imx}+e^{-imx})dx}{1+e^{2ix}}$, which almost resembles the example but not quite...
106102
How do I find the value of $$\int_{0}^{\infty} \frac{(\ln z)^2}{1+z^2}\mathrm{d}z$$ without using contour integration, - using the usual special functions, e.g., zeta/gamma/beta/etc.  Thank you,
91174
I was wondering how to determine $$\int x^2 e^{-x^2}\mathrm dx?$$ I tried to move the exponential into the integrator, and apply integration by parts, but that will make the integrand more complicated. Thank you!
120467
$$problem:\int_0^\infty cos(x^2)\ln(x)dx$$ $$Re\int_0^\infty e^{-ix^2}ln(x)dx$$ $$\Gamma(s)=\int_0^\infty x^{s-1}e^dx, \ \ \ \ x=u^2, dx=2udu$$ $$2\int_0^\infty u^{2s-1}e^{-u^2}du$$ $$\Gamma'(s)=4\int_0^\infty u^{2s-1}e^{-u^2}ln(u)du$$ $$\frac{\Gamma'(1/2)}{4}=\int_0^\infty e^{-u^2}ln(u)du$$ $$Re\int_0^\infty e^{-ix^2}ln(x)dx$$ $$Re\frac{1}{\sqrt{i}}\int e^{-u^2}ln(u)-\frac{1}{\sqrt{i}}\int e^{-u^3}ln(\sqrt{i})du$$ $$Re[\frac{\Gamma'(\frac12)}{4\sqrt{i}}-\frac{\sqrt{\pi}ln(i)}{4\sqrt{i}}]=Re[\frac{\Gamma'\frac(12)-\sqrt{pi}\ln(i)}{4\sqrt{i}}]$$ $$Re[\frac{\Gamma(\frac12)\psi(\frac12)-\sqrt{\pi}\ln(i)}{4\sqrt{i}}]$$ $$Re[\frac{\sqrt{pi}(-\gamma-2\ln(2))-\frac{\pi^{\frac32}}{2}i}{4\sqrt{i}}]$$ $$=\frac{\sqrt{2\pi}}{8}(-\gamma-2\ln(2))-\frac{\pi^{\frac32}}{16}$$
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.
42380
I have in trouble for evaluating following integral  $$\int_0^{\infty}  \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$$  It seems really easy, but I don't know how to handle it at all. (The results are well known, here I tried to evaluate it but I failed)  I tried to use the relation   $$\sqrt{1+x^{4}}-x^{2}=\frac{1}{\sqrt{1+x^{4}}+x^{2}}$$   but I couldn't find the desired results.
13977
I am trying to solve an integral in $k$-space where I want to calculate the ratio of the average energy of an arbitrary electronic band ($E(k)=k^n)$ to the square of average group velocity of electrons($v_g   = k^{n-1})$ approximately in the form of this expression given by $$L=\frac{\frac{\sum_{\mathbf{k}}k^{n}}{\sum_{\mathbf{k}}1}}{\left(\frac{\sum_{\mathbf{k}}k^{n-1}}{\sum_{\mathbf{k}}1}\right)^2}\;.$$  Now, I want to do this for $d$-dimensions, so I convert this sum to integral.  $$L=\frac{\dfrac{\int_{0}^{k_f}k^{n} d^d k}{\int_{0}^{k_f}1d^d k}}{\left(\dfrac{\int_{0}^{k_f}k^{n-1}d^d k}{\int_{0}^{k_f}1d^d k}\right)^2}$$ where $d^dk  = \dfrac{d \pi^{d/2}}{\Gamma(\frac{d}{2}+1)}k^{d-1}dk$ is the volume element in $d$-dimensional space.  I want to get a simple expression for $L$ and how it depends with $k_f,n$ and $d$. any physical insight into this integral could also be helpful.