New answers tagged improper-integrals
1
vote
Evaluate integral $\int_{-\infty}^\infty\Gamma(1+ix)\Gamma(1-ix)~dx$
$$I(a)=\int_{-\infty}^\infty\Gamma(1+iax)\Gamma(1-iax)\,dx$$
$$\Gamma(iax)\Gamma(1-iax)=\frac{\pi}{\sin(\pi iax)}$$
$$\Gamma(1+iax)\Gamma(1-iax)=a\pi\frac{x}{\sinh(\pi ax)}$$
$$I(a)=a\pi\int_{-\infty}^...
- 5,701
2
votes
Evaluating $\int_0^1 \frac{z \log ^2\left(\sqrt{z^2+1}-1\right)}{\sqrt{1-z^2}} \, dz$
Substitute $x^2=\sin 2\theta$ to remove square roots
\begin{align}
I=& \int_0^1 \frac{x \log ^2(\sqrt{1+x^2}-1)}{\sqrt{1-x^2}} \, dx\\
=& \ \frac12\int_0^{\pi/2}\ln^2(\cos\theta+\sin\theta-1)(\...
- 108k
1
vote
Integral $\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx$
Solution without special functions excluding the beta function and the gamma function
Start with a substitution, $x\rightarrow\frac{1-x}{1+x}$
$$I=\int_0^1\frac{\ln(1-x)}{\sqrt x\sqrt {1-x^2}}dx=\...
- 2,850
1
vote
Evaluate $\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}$
$$\boxed{\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^n}=\frac{n}{n^2-1} ; n>1}\tag1$$
Although a more generalized version would be for $m>n>0$,
$$\boxed{{\int_0^\infty \frac{x^{n-1}}{(x+\sqrt{1+...
- 5,701
1
vote
Evaluate $\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}$
Another method is to substitute $x+\sqrt{x^2+1}=t\implies \sqrt{x^2+1}-x=\frac1{t}\implies\mathrm dx=\frac{1+\frac1{t^2}}{2}\mathrm dt$:
$$\begin{align}\int_0^\infty\frac{\mathrm dx}{\left(x+\sqrt{x^2+...
- 4,468
3
votes
Sources about Glasser's master theorem
In the paper 'Which functions preserve Cauchy laws" Proceedings of the AMS 1977 Vol 37 pages 277- 286, yours truly proved the following : if $\mu$ is a singular bounded measure on $R$ then
$$f(x)=...
- 3,011
4
votes
Integral $\int_0^\infty \frac{|\sin\sqrt{qx}|-|\sin\sqrt{px}|}{x}dx$
A Frullani integral in disguise
\begin{align}
I=&\int_{0}^{\infty}
\frac{\left\vert\sin\sqrt{qx}\right\vert-
\left\vert\sin\sqrt{px}\right\vert}{x}\overset{\sqrt x\to x}{dx}\\
=&\
2 \int_{0}^{...
- 108k
1
vote
Integral evaluation of delta function with Gaussian
If $\delta$ is defined as a distribution, then the object (you wrote it as an integral) is ill-defined. This is because the product of distributions is not defined in general. Mathematica might be ...
- 183k
4
votes
Integral evaluation of delta function with Gaussian
The use of the symbol $\delta(x)$ by physicists is a calamity, since $ \delta$ is not a function but a measure which should be denoted by $\delta_a(dx)$ meaning $\int_R h(x)\delta_a(dx) =h(a)$ for a ...
- 3,011
3
votes
Integral evaluation of delta function with Gaussian
If you make the problem smoother and, since the Dirichlet kernel converges to a Dirac delta function, use
$$\delta(\tau)=\frac 2 \pi\,\,\underset{a\to \infty }{\text{limit }}\frac{\sin (a \tau )}{\...
- 274k
4
votes
Integrals of trignometric functions $\cot^{-1}\sqrt{1+\csc{\theta}}$ and $\csc^{-1}\sqrt{1+\cot{\theta}}$
For the first integral, substitute $\csc\theta=\frac{1+\sin^2t}{1-\sin^2t}$
\begin{align}\int_0^{{\pi}/{2}}\cot^{-1} &{\sqrt{1+\csc{\theta}}} \ {d\theta}
=\int_0^{\pi/2} \int_0^{\pi/4}\frac{\sqrt{...
- 108k
1
vote
Need help with $\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx$
\begin{align}
&\int_0^\infty\frac{\ln(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx\\
=& \int_0^1 \frac{\ln{(1+x)}}{1+x^2} dx- \int_0^1 \frac{x^3 \ln{x}}{(1+x^2)(1+x^3)}\overset{ibp}{dx}\\
=...
- 108k
0
votes
Proof of $\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$
Noting
$$ \mathcal{L}\{t\}(x)=\frac1{x^2},\mathcal{L}\{\sin^2(x)\}(t)=\frac{2}{t(t^2+4)} $$
one has
$$\int_0^{\infty} {\sin^2(x) \over x^2}\,dx=\int_0^{\infty} \sin^2(x)\mathcal{L}\{t\}(x) \,dx=\int_0^...
- 46.6k
0
votes
Proof of $\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$
Notice that $$\frac{\sin^2 x}{x^2}=\frac{1}{2}\Re\frac{1-e^{2iz}}{z^2}:=\Re f(z).$$
Thus, we try to do the integral $$\int_{C+D+A+B:=\gamma}f(z)dz,$$ where the contour is
Let's first consider $\int_C ...
- 2,564
5
votes
Accepted
Alternating series of Laplace transforms is zero implies that Laplace transform is zero?
Yes, if that summation is exactly zero for all $r$, then we shall have $F=0$ being the zero function. To prove this, it helps to consider the alternating summation as another sort of transform between ...
- 4,210
3
votes
Accepted
Evaluating $\int_{0}^{\infty}e^{-\mu\lvert u-s \rvert-\lambda s}ds$
When you integrate $\int e^{ks}ds = \frac1ke^{ks}+c$, that is how you get the factors on the denominator, i.e.
\begin{align*}
\int_0^\infty e^{-\mu|u-s|-\lambda s}ds
&= e^{-\mu u}\int_0^u e^{(\mu ...
- 1,073
1
vote
Compute: $\int_0^{\infty}e^{-\pi x}\text{Ei}(x)dx$
OMG, nevermind, I just realized something. Notice, by definition of Ei(x),
$$\frac{1}{a}\lim_{x\rightarrow0}(\text{Ei}((a-1)x)-\text{Ei}(x))=\frac{1}{a}\int_{0}^{\infty}\frac{e^{-(a-1)x}-e^{-x}}{x}dx$$...
- 1,501
1
vote
Integral: $\int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx$
Let $s_{\pm}=\sqrt{p^2+1}\pm p$ and note that
$$\int_0^\infty \sin bx \ \tan^{-1}\frac{2ax}{x^2+c^2}\ dx
\overset{x\to c x}=c I(\frac ac, bc)
$$
where
\begin{align}
I(p,q)=& \int_0^\infty \sin qx \...
- 108k
0
votes
Seemingly Do-able Gaussian
$$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\exp \left(\sqrt{x^2+y^2}\right) \exp \left(-\frac{1}{2} \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)\right)dydx=\\\int _0^{\infty }\int _{-\pi }^{...
- 5,058
-1
votes
Seemingly Do-able Gaussian
Define
$$F(a,b)=\int_{\Bbb R^2}\exp\left(x^2+y^2\right)\exp\left(-\frac{1}{2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)\right)\mathrm dx\mathrm dy$$
Everything is separable,
$$F(a,b)=\int_{y=-\infty}...
- 12.9k
0
votes
Juantheron-like integral
$$\int_0^{\infty } \frac{1}{\left(1+x^2\right) (1+\tan (x))} \, dx=\\\int_0^{\infty } \left(\mathcal{L}_x\left[\frac{1}{1+\tan (x)}\right](t)\right) \left(\mathcal{L}_x^{-1}\left[\frac{1}{1+x^2}\right]...
- 5,058
0
votes
Proving the sum of two independent Cauchy Random Variables is Cauchy
I will try to show a detailed proof of the above $(4)$ to $(5)$ (for those who does not understand of course). I am not able to comment the above reply so I will reply from here.
$$J(m)=\int_{0}^\...
- 33
2
votes
Generalizing $\int_{0}^{\infty} \frac{1}{( 1+x^2) ( 1+(2+x)^2) }\text{d}x+\int_{0}^{1}\frac1{(1+x^2)( 1+(2-x)^2)}\text{d}x=\frac{\pi}{8} .$
Too long for a comment
You have been very lucky to use $a=2$ for the more general case of
$$A=\int_{0}^{\infty} \frac{dx}{( 1+x^2) ( 1+(a+x)^2) }+\int_{0}^{1}\frac{dx}{(1+x^2)( 1+(a-x)^2)}$$
The ...
- 274k
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