Solve this integral$$\int_{0}^{\pi}{\log(\sin x)}\,\mathrm dx$$   I have tried this by breaking the limit from $0$ to $\pi/2$ and $\pi/2$ to $\pi$, but I am unable to solve the 2nd part.
121569
From this paper:     Shouldn't $du$ be $dt$? And also how do you get from that line to the final result if $du$ is not $dt$?
65916
I stumbled upon this integral $$I=\int_1^3 \frac{\log x}{x^2+3}dx$$ Since the answer given by Wolfram-Alpha is $I=\frac{\pi \log 3}{12\sqrt3}$ I tried to work near the bounds of the integral because as a indefinite integral it will involve some polylogarythms which I want to avoid. But not a single substitution  like $x=\frac{1}{y}$ or $x-2=u$ gave any better look to this integral. And when integrated by parts I feel like it got worse. Could you perhaps help me on this?
122105
Possible Duplicate:          $$\int_{0}^{1} dx\frac{\log(1+x)}{1 + x^2}$$  I am having a hard time deriving the answer, $\frac{\pi}{8} \log(2) $.  I have tried Taylor expansion of both numerator and denominator, both seem too complicated and fruitless.
28772
How to solve the following integral.  $I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$ where $|\theta|&lt;1$
2754
Show that,$$\int_0^\pi \bigg|\dfrac{\sin nx}{x}\bigg|\mathrm{d}x \ge \dfrac{2}{\pi}\bigg(1+\dfrac12+\cdots+\dfrac{1}{n}\bigg)$$    I could not approach the problem at all. Please help.
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.
85817
How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.
68185
Please integrate $\frac{1}{\log(x)}$ and illustrate the steps of your method of integration. I have already tried integration by parts following the ILATE rule and otherwise. Eventually it forms a loop that takes me back to the direct or substituted term where I started the process of integration.
57523
How can I evaluate the following integral?     $$ \int_0^{\pi/2} \log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\, \frac{1}{\cos\left(x\right)}\,{\rm d}x\,, \qquad\left\vert\,a\,\right\vert \le 1$$   I tried differentiating under the integral with respect to the parameter $a$, and I also tried expanding the log term in a Taylor series and then switching the order of integration and summation. I ran into difficulties with both approaches.
75595
I have been attempting to solve this integral for several days in preparation for an exam and keep reaching dead ends.  I(a,b) = $\int_0^{\pi/2} \ln(a^{2}\cos^2(x)+b^{2}\sin^2(x)) dx$, where $a,b&gt;0$  My best approach so far is defining:  $F(x) = I(x, b)$  Then, I know:  $F(b) = I(b, b) = \pi/2 * \ln(b^2)$  Then I am now attempting to find $F'(x)$ using differentiation under the integral sign, and integrating from b to a.  This is where I am stuck - no matter what I do, I cannot solve the integral. Am I in the right direction or should this be solved in a different manner? I also considered something elliptical due to the acos, bsin in the original integral, yet it has also reached a dead end.
32720
$${\dfrac{1}{\pi^2}}\int_{0}^{\infty}\dfrac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$$   I tried substituting $1/x$ for $x$ but the the only change in the integral is that the $\sqrt{x}$ moves in the numerator from the denominator. I don't understand what to substitute. $\tan{x}$ doesn't seem to work.
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!
12062
I have no idea where to even start. WolframAlpha cant compute it either.  $$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$  I think it can be done with series, but I am not sure, can someone help a little so I can get a start??  Thanks!
120781
I ran across this integral I get no where with. Can someone suggest a method of attack?.  $$\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}$$  I tried series, imaginary parts, and so forth, but have made no progress.    Thanks very much.
47607
I'm trying to find $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx$ where $\alpha&gt;0$ is real.  My approach was to take an integral along the real line from $1/R$ to $R$, around the circle counterclockwise to $-R$, along the real line to $-1/R$, and then around the circle clockwise to $1/R$.  I have encountered 2 problems with this:   This path encloses one pole, at $z=\alpha i$.  I found the residue at $z=\alpha i$ to be $\frac{\ln(\alpha)+i\pi/2}{2\alpha i}$.  However, this gives me that $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx=\frac{\pi(\ln(\alpha)+i\pi/2)}{2\alpha}$.  Since I have a real function integrated over the real line, there cannot be an imaginary part.  Where did I go wrong?  (Also, doing a few examples, the correct answer seems to be $\frac{\pi\ln(\alpha)}{2\alpha}$, the same as I have but without the imaginary part.) At first chose my path so instead of going all the way around the upper semicircle, it only went 3/4 of the way around, as I wanted to avoid anything that might go wrong with the discontinuity of $\log(x)$ at the negative real axis.  When I do this, though, I get a different answer than before (the denominator of the fraction is $\alpha(1-e^{3\pi i/4})$ instead of $2\alpha$.  What am I doing wrong that gives me different answers?
76201
I need help solving this integral: for which values of the parameter $p$ is this improper integral convergent?  $$\int_{0}^{\pi/2}\tan^{p}(x) \; dx$$  Thanks a lot!
59067
Is the integral $$\int_1^\infty\frac{x^{-a} - x^{-b}}{\log(x)}\,dx$$ convergent, where $b&gt;a&gt;1$?  I think the answer lies in defining a double integral with $yx^{(-y-1)}$ and applying Tonelli's Theorem, but the integral of $\frac{x^{-a}}{\log x}$ is still not integrable. Any ideas?