2
votes
0answers
64 views

Contour Integral $ \int_{0}^1 \frac{\ln{x}}{\sqrt{1-x^2}} \mathrm dx$

I need help evaluating this with contour integration$$ \int_{0}^1 \frac{\ln{x}}{\sqrt{1-x^2}} \mathrm dx$$ I am not sure as to how to work with the branch cuts of both $\ln{x}$ and $\sqrt{1-x^2}$ ...
5
votes
1answer
103 views

Prove $\int_0^1 \frac{\ln(1+t^{4+\sqrt{15}})}{1+t}\mathrm dt= -\frac{\pi^2}{12}(\sqrt{15}-2)+\ln (2) \ln(\sqrt{3}+\sqrt{5})+\ln(\phi) \ln(2+\sqrt{3})$

Prove that: \begin{equation} \int_0^1 \frac{\ln\left(1+t^{4+\sqrt{15}}\right)}{1+t}\mathrm dt= -\frac{\pi^2}{12}(\sqrt{15}-2)+\ln (2) \ln(\sqrt{3}+\sqrt{5})+\ln(\phi) \ln(2+\sqrt{3}) ...
2
votes
1answer
23 views

Showing that $\tan(\pi z) = z$ has exactly three solutions in the strip $|\Re(z)| < 1$

We can't use Rouche's theorem here directly, so we have to apply the argument principle. If $f(z) = \tan(\pi z) - z$ , then $f'(z) = \pi \sec^2(\pi z) - 1$. Choose the rectangle $\Gamma$ with ...
0
votes
0answers
31 views

What advanced methods in contour integration are there?

It is well known how to evaluate a definite integral like $$ \int_{0}^\infty dx\, R(x), $$ where $R$ is a rational function, using contour integration around a semicircle or a keyhole. Most complex ...
2
votes
1answer
28 views

Complex Green's Theorem

I want to integrate $\int_{\partial R} |e^{zt}|dz$ where $R\subseteq \mathbb{C}$ is a rectangle whose sides are parallel to the coordinate axes. I want to use a complex version of green's theorem, but ...
6
votes
2answers
145 views

A strange answer for $\int _{-1}^1 \log x\; dx$

I typed $\int _{-1}^1 \log x\; dx$ on Wolfram Alpha. It is giving the answer to be $-2+i\pi$. Can someone please explain what is happening?
1
vote
3answers
84 views

Guidance or advice with $I=\int_0^{2\pi}\frac{1}{4+\cos t}dt$

Let $$ \begin{align} I=\int_0^{2\pi}\frac{1}{4+\cos t}dt \end{align} $$ I would like to evaluate this integral using cauchhy's Integral formula, I understand that I have to convert this into a form ...
10
votes
0answers
193 views

The closed form of $\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$

What tools, ways would you propose for getting the closed form of this integral? $$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
4
votes
0answers
140 views

${\mathfrak{I}} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$ and $\int_{0}^{\pi/2} \frac{\log \cos x}{x^2}\:\mathrm{d}x$

I have found the following new result connecting two rational log-cosine integrals. Proposition. \begin{align} \displaystyle & {\mathfrak{I}} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos ...
2
votes
0answers
52 views

Choose appropriate contour for a complex integral

I have a problem to solve integral $$ I = \int^{\infty}_0 \frac{\mathrm{d}x}{(x-z)(1+x^2)^{\kappa+2}} $$ I can solve the same integral with borders $-\infty$ to $\infty$ using residue theorem but ...
8
votes
1answer
262 views

Evaluation of $\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$

I need some hints, clues for getting the closed form of $$\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$$
0
votes
0answers
92 views

On an application of the Abel-Plana formula

Referring to a previous question, i am having a hard time trying to do the integral: $$f(s)=-i\int_{0}^{\infty}\frac{\log \left[1+\frac{\left(s\log(1+ix) \right )^{2}}{4\pi ^{2}} \right ]-\log ...
2
votes
0answers
60 views

Evaluate $\int_0^\infty x^{\lambda-1} \exp\left(-ax-b\sqrt x-\frac{c}{\sqrt x} - \frac{d}{x}\right) \: dx$

Is there a closed form for the integral $$\int_0^\infty x^{\lambda-1} \exp\left(-ax-b\sqrt x-\frac{c}{\sqrt x} - \frac{d}{x}\right) \: dx?$$ where $\lambda>0$, $a>0$, $d>0$ and where $b$, ...
2
votes
0answers
36 views

Conditions for changing the order of integration for contour integral.

I assume an integral $$I=\int_0^\infty f(x)g(x)\mathrm dx \tag{1}$$ where the function $f(x)$ can be represented as a contour integral in complex plane: $$f(x)=\oint_\Delta ...
0
votes
0answers
60 views

Integral $\int^\infty_{-\infty}\int^\infty_{-\infty}(\frac{(x-x_1)^2+(y-y_1)^2}{s_1^2}+1)^{-a_1-1}(\frac{(x-x_2)^2+(y-y_2)^2}{s_2^2}+1)^{-a_2-1}dxdy$

Under $x_i,y_i\in\mathbb R$, $s_i>0$ and $a_i>0$ for $i=1,2$, is there any good function to express the following integral? $$\int^\infty_{-\infty}\int^\infty_{-\infty} ...
1
vote
2answers
41 views

Definite integrals with complex analysis

Can anyone explain me why: $$\int_{C_r}\frac{1}{z} \mathrm{d}z+\int_{C_r}\frac{e^{iz}-1}{z}\mathrm{d}z\stackrel{r \to 0^+}{=}\pi i$$ $C_r$ is half circle from $r$ to $-r$
3
votes
2answers
163 views

Integration of exponential and square root function

I need to solve this $$\int_{-\pi}^{\pi} \frac{e^{ixn}}{\sqrt{x^2+a^2}}\,dx,$$ where $i^2=-1$ and $a$ is a constant.
2
votes
2answers
64 views

Cauchy distribution characteristic function

I know that it's easy to calculate integral $\displaystyle\int_{-\infty}^{\infty}\frac{e^{itx}}{\pi(1+x^2)}dx$ using residue theorem. Is there any other way to calculate this integral (for someone who ...
1
vote
1answer
102 views

Complex Integral Question

I'm trying to evaluate the following integral, in preparation for my exam tomorrow; $$\int_{0}^{\infty} \frac{\cos(2x) - 1}{x^2} dx$$ However, I'm having a lot of issues with it. I was initially ...
2
votes
1answer
133 views

What is ${\mathfrak{R}} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$?

This is a new integral that I propose to evaluate in closed form: $$ {\mathfrak{R}} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$$ where $\log (z)$ denotes the principal value of ...
14
votes
3answers
307 views

The 3 Integral $\int_0^\infty {x\,{\rm d}x\over \sqrt[3]{\,\left(e^{3x}-1\right)^2\,}}=\frac{\pi}{3\sqrt 3}\big(\log 3-\frac{\pi}{3\sqrt 3} \big)$

Hi I am trying evaluate this integral and obtain the closed form:$$ I:=\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}=\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3} \right). $$ The ...
9
votes
1answer
367 views

Help with a troublesome double integral

I'm having difficulty with a double integral $$-2i\int_{0}^{\infty}\int_{0}^{\infty}\frac{dxdt}{t(e^{2\pi x}-1)(e^{2\pi t/s}-1)}\left[\cos(t\log(1-ix))-\cos(t\log(1+ix))\right]$$ where ...
13
votes
2answers
438 views

Integral $\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx$

I am having trouble showing this equality is true$$ \int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma ...
5
votes
2answers
127 views

Integral $\int_0^{\pi/4}\frac{x^2\tan x}{\cos^2 x}dx=\frac{\log 2}{2}-\frac{\pi}{4}+\frac{\pi^2}{16}$

Hi I am trying to evaluate the definite integral which has a closed form given by: $$ \mathcal{I}=\int_0^{\pi/4}\frac{x^2\tan x}{\cos^2 x}dx=\frac{\log 2}{2}-\frac{\pi}{4}+\frac{\pi^2}{16}. $$ We can ...
1
vote
2answers
100 views

Is there always an alternative way to compute integral other than complex integral?

Sometimes we have to compute integrals that are not easy to calculate so that we need to depend on the method of complex integrals like the residue method. But I became curious about possibility of ...
6
votes
3answers
162 views

Integral $\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\frac{17\pi^4}{360}$

Hi I am trying to integrate $$ \mathcal{I}:=\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\int_0^1\int_0^1 \frac{\log(1-x)\log(1-y)}{1-xy}dx \,dy $$ A closed form does ...
0
votes
1answer
44 views

Using Poisson's integral formula

The problem asks to prove the following equality using Poisson's integral formula (or Poisson kernel, if I understood correctly from Wikipedia): $$\int_0^{2\pi} \frac{e^{\cos ...
5
votes
4answers
210 views

Integral $\int_0^1\frac{dx}{\sqrt{\log \frac{1}{x}}}=\sqrt \pi$

Hi I am trying to prove this result below $$ \mathcal{J}:=\int_0^1\frac{dx}{\sqrt{\log \frac{1}{x}}}=\sqrt \pi. $$ The result is quite interesting however I realized I am not familiar with working ...
7
votes
3answers
193 views

Prove that $\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x$ equal $\log 2$

This integral poppep up recently $$\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x = \log 2$$ A solution using both real and complex analysis is welcome. I tried ...
11
votes
4answers
185 views

Integral $\int_0^\infty \log(1-e^{-a x})\cos (bx)\, dx=\frac{a}{2b^2}-\frac{\pi}{2b}\coth \frac{\pi b}{a}$

$$\mathcal{J}:=\int_0^\infty \log(1-e^{-a x})\cos (bx)\, dx=\frac{a}{2b^2}-\frac{\pi}{2b}\coth \frac{\pi b}{a},\qquad \mathcal{Re}(a)>0, b>0. $$ I tried to write $$ \mathcal{J}=-\int_0^\infty ...
10
votes
1answer
157 views

Integral $\int_0^\infty \frac{\cos x}{x}\left(\int_0^x \frac{\sin t}{t}dt\right)^2dx=-\frac{7}{6}\zeta_3$

Hi I am trying to prove this below. $$ I:=\int_0^\infty \frac{\cos x}{x}\left(\int_0^x \frac{\sin t}{t}dt\right)^2dx=-\frac{7}{6}\zeta_3 $$ where $$ \zeta_3=\sum_{n=1}^\infty \frac{1}{n^3}. $$ I am ...
7
votes
3answers
181 views

Integral $\int_0^{\pi/4}\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}=\frac{\pi}{8a}$

I am trying to prove this interesting integral$$ \mathcal{I}:=\int_0^{\pi/4}\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}=\frac{\pi}{8a},\qquad \mathcal{Re}(a)\neq 0. $$ This result is breath taking but I ...
2
votes
2answers
151 views

Integral $\int_0^\infty \frac{\sin a x-\sin b x}{\cosh \beta x}\frac{dx}{x}$

Hi I am trying to prove this interesting integral $$ \mathcal{I}:=\int_0^\infty \frac{\sin a x-\sin b x}{\cosh \beta ...
4
votes
1answer
187 views

$\int_0^\infty \frac{x\cos ax}{1+x^2}\coth \frac{\pi x}{4} dx=\frac{\pi}{2}e^{-a}+\cosh a\log \coth \frac{a}{2}+2\sinh a \arctan(e^{-a})-2$

Hi I am trying to prove this $$ \int_0^\infty \frac{x\cos ax}{1+x^2}\coth \frac{\pi x}{4} dx=\frac{\pi}{2}e^{-a}+\cosh a\log \coth \frac{a}{2}+2\sinh a \arctan(e^{-a})-2,\qquad a>0. $$ What a ...
3
votes
2answers
129 views

$\int_0^\infty \frac{\cos a x-\cos b x}{\sinh \beta x}\frac{dx}{x}=\log\big( \frac{\cosh \frac{b\pi}{2 \beta}}{\cosh \frac{a\pi}{2\beta}}\big)$

Hi I am trying to prove this interesting integral $$ \mathcal{I}:=\int_0^\infty \frac{\cos a x-\cos b x}{\sinh \beta x}\frac{dx}{x}=\log\left( \frac{\cosh \frac{b\pi}{2 \beta}}{\cosh ...
2
votes
3answers
163 views

Integral $\int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}$

EDIT: Small Typo Fixed now, Thanks to Sir Chen Wang! Hi I am trying to prove this result without using a series approach $$ \int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}. ...
4
votes
3answers
74 views

Integral $I(a,b)= P\int_{0}^{\pi}\frac{d\theta}{a-b\cos\theta}$

Hi I am trying to calculate this integral $$ I(a,b)= P\int_{0}^{\pi}\frac{d\theta}{a-b\cos\theta},\quad 0 <a<b,\quad a,b\in \mathbb{R}. $$ We can first write $$ I(a,b)=\frac{1}{2} ...
2
votes
1answer
63 views

find the value of the following (real) integral

$$\int_0^{+\infty} \frac{x^2}{(x^2+9)(x^2+4)} \, dx = \int_0^{+\infty}f(x) \, dx$$ Here is my attempt, if you could please look over it and tell me if it's correct or not: First of all, the ...
1
vote
1answer
42 views

Integral Continuation $\Gamma(z)=\int_{0}^{1} e^{-t} t^{z-1} dt +\int_{1}^{\infty} e^{-t} t^{z-1}dt$

I am trying to obtain an analytical continuation for $\Gamma(z)$ into the region of the complex plane characterized by $\Re(z) \leq 0$ but am stuck. Starting from the integral definition of ...
4
votes
4answers
162 views

Integral $P\int_0^\infty \frac{x^{\lambda-1}}{1-x} dx$

I am trying to calculate the following principle value integral \begin{equation} P\int_0^\infty \frac{x^{\lambda-1}}{1-x} dx \end{equation} for $\lambda \in [0,1].$ I tried to turn this into a ...
17
votes
7answers
361 views

Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$

Hi how can we prove this integral below? $$ I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3} $$ I tried to use $$ I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx $$ and now ...
4
votes
1answer
111 views

Integral $\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}$

I am trying to prove this interesting integral $$ I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}. $$ I tried using $y=1+x^3$ but that didn't ...
6
votes
1answer
125 views

$\int_0^\infty \log(1+x^2) \frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx=4\sqrt 2-\frac{16}{\pi}+\frac{8\sqrt 2}{\pi}\log(\sqrt 2+1)$

I am trying to prove this interesting integral $$ I:=\int_0^\infty \log(1+x^2) \frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx=4\sqrt 2-\frac{16}{\pi}+\frac{8\sqrt 2}{\pi}\log(\sqrt 2+1). $$ I ...
1
vote
3answers
138 views

Integral $\int_0^\infty \frac{\sin^2 ax}{x(1-e^x)}dx=\frac{1}{4}\log\left( \frac{2a\pi}{\sinh 2a\pi}\right)$

How can we prove this ${\it{interesting}}$ integral $$ I:=\int_0^\infty \frac{\sin^2 ax}{x(1-e^x)}dx=\frac{1}{4}\log\left( \frac{2a\pi}{\sinh 2a\pi}\right) $$ I to write $$ I=\frac{1}{2}\int_0^\infty ...
5
votes
1answer
111 views

A solution for $\int^{2\pi}_0e^{\cos \theta}\cos(a\theta -\sin \theta)\,d \theta $

It can be proved using complex analysis that $$\tag{1}\int^{2\pi}_0e^{\cos \theta}\cos(n\theta -\sin \theta)=\frac{2\pi}{n!}$$ My initial thought that, we use the Gamma function for non-integer ...
1
vote
0answers
57 views

What will be the integration region?

Where $\Omega_s$ is new integration region, due to change in geometry integration region will also change. Also note that $\Omega_l$ is $\Omega$ with $|x|<l$ is the integration region for ...
7
votes
5answers
195 views

Simple Integral $\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}$.

Hellow I am trying to prove this result. $$ I:=\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}. $$ The indefinite integral exists for this integral. The function $\cot^{-1} x$ is the arc-cotangent ...
3
votes
3answers
207 views

Integral $\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx=-\frac{\pi}{4\alpha}\text{arcsinh}\alpha$

Hi I am trying to prove this $$ I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\, \frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x ...
13
votes
6answers
343 views

Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$

Hi I am trying to solve this integral $$ I:=\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a,\qquad |a|\leq 1. $$ It gives beautiful result for $a=1$ $$ \int_0^1 \log ...
5
votes
1answer
92 views

Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x^2+1} dx$

I'm trying hard to solve this integral but I still don't know how... $$\int_{0}^{\infty} \frac{\sin{x}}{x^2+1} dx$$ The integral from $-\infty$ to $\infty$ is quite easy, but how could we integrate ...