Linked Questions

12
votes
4answers
3k views

Integral:$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $

How to evaluate: $$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $$ Maybe we can evaluate it using the well-known result:$\int_{0}^{\frac{\pi}{2}} \ln{\sin t} \text{d}t=\int_{0}^{\frac{\pi}{2}...
0
votes
0answers
59 views

How to evaluate $\int_{0}^{\infty} \frac{\log(x^{2}+1)}{x^{2}+1}$ [duplicate]

I tried to find $f(a) = \int_{0}^{\infty} \frac{\log(x^{2}+a^{2})}{x^{2}+b^{2}}$. After differentiating I get : $f(a) = \frac{\pi \log(a+b)}{b} + C$. But it's not easy to find this constant. I ...
4
votes
2answers
281 views

Find $\lambda$ if $\int^{\infty}_0 \frac{\log(1+x^2)}{(1+x^2)}dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}dx$

Problem : If $\displaystyle\int^\infty_0 \frac{\log(1+x^2)}{(1+x^2)}\,dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}\,dx$ then find the value of $\lambda$. I am not getting any clue how to proceed ...
4
votes
2answers
187 views

Improper integral of $\frac{\log\left(\sqrt{x^2+a^2}\right)}{x^2+b^2}$

Show that $$\int_{-\infty}^\infty \frac{\log\left(\sqrt{x^2+a^2}\right)}{x^2+b^2}\,dx = \frac{\pi}{b}\log\left(a+b\right)$$ for $a,b>0\in\mathbb{R}$. I stumbled on this answer empirically, but I'm ...
1
vote
1answer
140 views

Solution of $\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$ [closed]

How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.
3
votes
3answers
567 views

Evaluate an improper integral involving log [duplicate]

Studying for complex analysis, I stumbled upon this problem. Evaluate $$\int_{0}^{\infty} \frac{\log(1+x^2)}{1+x^2}~dx.$$ So the integrand suggests that I need to use the function $f(z)=\frac{\log 1+z^...
14
votes
3answers
2k views

Can $ \int_0^{\pi/2} \ln ( \sin(x)) \; dx$ be evaluated with “complex method”?

Can the following integral be evaluated using complex method by substituting $\sin(x) = {e^{ix}-e^{-ix} \over 2i}$? $$ I=\int_0^{\pi/2} \ln ( \sin(x)) \; dx = - {\pi \ln(2) \over 2}$$
2
votes
1answer
73 views

Big trouble in evaluating an Improper integral in complex analysis [duplicate]

I am having trouble in evaluating $\int_{0}^\infty \left(\frac{log(x^2+1)}{x^2+1}\right)dx$ I proceeded by taking $\int_{C}\left(\frac{log(z+i)}{z^2+1}\right)dz$ I don't know how to draw contour here ...
6
votes
2answers
317 views

Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$

Prove that $$\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx=\frac{\pi\sqrt{2}}{2}\log\left(1+\frac{\sqrt{2}}{2}\right).$$ I managed to prove this result with some ...
6
votes
1answer
230 views

Evaluate $\int_{-\infty}^{\infty} \frac{\log(1+x^2) dx}{1+x^2}$ Using Complex Analysis

Evaluate: $$\int_{-\infty}^{\infty} \frac{\log(1+x^2) dx}{1+x^2}$$ Using complex analysis, contour integration. This function has no poles at all. Try the contour $C$ Obviously, $$\oint_{C} ...
3
votes
3answers
326 views
18
votes
4answers
626 views

Integrate $ \int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2$

I am trying to evaluate this integral. $$ I=\int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2 $$ Note $$ \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}, \ |x| < 1. $$ I was trying to do ...
1
vote
1answer
472 views

Complex integration Question - Contour Method [duplicate]

I'm asked to find: $$\int_{-\infty}^\infty \frac{\ln(x^2+1)}{1+x^2} dx $$ Attempt Considering $$ \oint \frac{\ln(z^2+1)}{(z+i)(z-i)} dz $$ So first I find the branch points of the function. This ...
6
votes
5answers
301 views

Find the following integral (most likely substitution)

$$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$$ I tried letting $x^2=\tan \theta$ but it didn't work. What should I do? Please don't give full solution, just a hint and I will continue.
3
votes
0answers
74 views

Typical Contour Inegral Proof

I'm trying to prove that the following contour integral approaches 0 as R -> $\infty$. How exactly would we go about doing this? $$ \int{\log\left(z^{2} + 1\right) \over 1 + z^{2}}\,{\rm d}z\quad \...

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