$ b&gt;a&gt;0$. Calculate the integral $\displaystyle \int_{0}^{\infty} \frac{e^{-ax}-e^{-bx}}{x}\ dx $  Integration by parts doesn't work, nor can I find a proper substituion.
27738
Prove that      $$ \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right) $$   My Attempt:  Define the function $I(a,b)$ as  $$ I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx $$  Differentiate both side with respect to $a$ to get  $$ \begin{align} \frac{dI(a,b)}{da} &amp;= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &amp;= \int_{0}^{\infty}e^{-ax}\,dx\\ &amp;= -\frac{1}{a}(0-1)\\ &amp;= \frac{1}{a} \end{align} $$  How can I complete the proof from here?
14940
$$\int \frac{\mathrm{d}x}{x}$$ If I integrate this by parts ($\displaystyle u=\frac{1}{x}, \mathrm{d}u = -\frac{\mathrm{d}x}{x^2}, \mathrm{d}v= \mathrm{d}x, v = x$), then why does this happen?  $$\int \frac{1}{x} \: \mathrm{d}x = u v - \int v \:\mathrm{d}u = \frac{x}{x} - \int x\left(\frac{-1}{x^2}\right) \:\mathrm{d}x=1+\int \frac{1}{x}\:\mathrm{d}x$$  So $\displaystyle\int \frac{1}{x}\:\mathrm{d}x = 1 + \int \frac{1}{x} \:\mathrm{d}x$ and cancel out both sides of $\displaystyle\int \frac{1}{x} \:\mathrm{d}x$ and I get $0=1$. How can I find the error in the process because right now I cant see it.
8650
I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
48568
Evaluate $$\int_0^{+\infty}\frac{\log x}{1+e^x}\,dx.$$  I have tried using Feynman's Trick (in several ways, but for example by introducing a variable $a$ such that $I(a)=\int_0^{+\infty}\frac{\log ax}{1+e^x}\,dx$), but that doesn't seem to work. Also integration by parts and all kinds of substitutions make things worse (I have no idea how to substitute such that $\log$ and $\exp$ both become simpler.  (Source:  - Level 5/5)
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
22397
I want to compute this integral $$\displaystyle\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$$  What I did was the following. I substituted $x=t^{6}$, so that my $dx= 6t^{5} \ dt$ and so the integral changes to $$\int_{0}^{\infty} \frac{t^2}{1+t^{12}} \cdot 6 t^{5} \ dt =6 \cdot \int_{0}^{\infty} \frac{t^7}{1+t^{12}} \: dt$$  Now If I substitute $t^{4}=v$ then what I will be having is the following integral $$\frac{6}{4}\cdot \int_{0}^{\infty} \frac{v}{1+v^{3}} \ dv$$  Now I can write $1+v^{3} = (1+v) \cdot (1-v+v^{2})$ and so I have   \begin{align*} \int_{0}^{\infty} \frac{v}{1+v^{3}}\: dv &amp;= \int_{0}^{\infty}\biggl[\frac{1}{3}\cdot \frac{v+1}{1-v+v^{2}} - \frac{1}{3} \cdot \frac{1}{1+v}\biggr]\: dv \end{align*}  Now the point is that the integral of $1/(1+v) \to \infty$, so I am not sure if this is the right way to do. Can anyone suggest anything?
8145
I have been self-studying CA and find it very interesting. So, working through problems in a book I have, I ran across  $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{2\pi x}-1}dx=\frac{1}{4}\coth(a/2)-\frac{1}{2a}$$  and $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{x}+1}dx=\frac{1}{2a}-\frac{\pi}{2\sinh(\pi a)}$$  For the former, I wrote it as $\frac{e^{aiz}}{e^{2\pi z}-1}$ and used a rectangular contour with vertices $0, \;\ R. \;\ R+i, \;\ i$  $e^{2\pi z}-1$ has poles at $ni$. Of these, I think $i$ only lies within the contour.   Unless I am in error, I then calculated the residue at $i$ to be $\frac{e^{-a}}{2\pi}$  So, $2\pi i(\frac{e^{-a}}{2\pi})=ie^{-a}$  Now, where I get hung up is setting up the integrals around the contour. Two of which should tend to 0 as $R\to \infty$.  Here is what I done.  $$\int_{0}^{R}\frac{e^{iax}}{e^{2\pi x}-1}dx+\int_{0}^{\infty}\frac{e^{ai(R+iy)}}{e^{2\pi (R+iy)}-1}idy+\int_{R}^{0}\frac{e^{ai(x+i)}}{e^{2\pi (x+i)}-1}dx+\int_{\infty}^{0}\frac{e^{ai(iy)}}{e^{2\pi iy}-1}dy=-\sinh(a)$$  I am unsure of the limits on the second and fourth integrals.  I am not so sure this is correct. The second and fourth ones, which represent the vertical sides,  should tend to 0 as $R\to\infty$. I hope :).   This than gave me $(1-e^{-a})\int\frac{e^{iax}}{e^{2\pi x}-1}dx=-\sinh(a)$.  Which does not look correct. I did manage to solve this using series and $\pi csc(\pi z)$, but the contour I am unsure of.   Can someone lend a hand here?. Any advice on either would be appreciated.   For the other one, I only posted it because it looks similar, but I believe is actually more involved and 'tougher' if you will. If I can get one, perhaps I can manage to evaluate the other.  With that one, I think the same rectangle with the same vertices can be used, but $\pi i$ would have to be avoided. Perhaps a Principal Value in there somewhere. But, I was told to use vertices $0, \;\ R, \;\ R+2\pi i, \;\ 2\pi i$  Since I am relatively new to CA, setting up those integrals is the confusing part.    I can do easier types of contour integrals, but want to learn more about these more challenging ones.   Thanks for any assistance.
31532
$\displaystyle \int_{0}^{\frac{\pi}{2}} \ln(\sin(x))dx=-\frac{\pi}{2}\ln(2)$ is an integral that is common.  But, how can we show $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx=\frac{{\pi}^{3}}{24}+\frac{\pi}{2}\ln^{2}(2)$?.  Does anyone have any ideas on how to approach $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$?.  Thank you very much.
61051
I need to evaluate the following function and then check my answer by taking the derivative: $$\int e^{ax}\cos(bx)\,dx$$ where $a$ is any real number and $b$ is any positive real number.    I know that you set $u=\cos(bx)$  and $dv=e^{ax} dx$, and the second time you need to integrate again you set $u=\sin(bx)$ and $dv=e^{ax}dx$ again.  It eventually simplifies down to $$\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$  Now I know to move the integral on the left side to the right side so that I can just divide by the constant to solve.  Here is my problem:  I know I need to solve the right side to be:  $$\frac{e^{ax}\left(a\cos(bx) + b\sin(bx)\right)}{a^2+b^2} + C.$$  To divide by the constant, I multiplied everything on the right side by  $$\frac{a^2}{b^2+1}.$$ but this leads me to get $b^2 + 1$ on the bottom instead of $a^2 + b^2$.  I will show what I am doing in detail:  After setting the initial integral equal to: $$\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right) + C$$ I simplify: $$\int e^{ax}\cos(bx)\,dx = \frac{a^2}{b^2+1}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx)\right)\right) + C$$  If this is already wrong, can someone point be in the right direction?    If I have not gone wrong yet, I can edit to show the rest of my work.
68926
I want to find the following integral: $$\int_{0}^{\infty} \frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)} dx$$ I have been trying to apply simple substitutions like $u=e^x$, to simplify the integral without any luck. I simplified it to  $$\int_{0}^{\infty} \frac{1}{x(e^x+1)}-\frac{1}{x(e^{3x}+1)}dx$$  which seems to have increased my problems.  Wolfram Alpha gives the answer to be 0.5493... and the graph looks like a Gaussian distribution.  Somehow, I feel that this integral is very non-trivial due to the 'troublesome' $x$ in the denominator.
18956
I have a problem to solve and I have something that i don't know how to do.  The half-space Gaussian integral is given :  $$\int_{0}^\infty \exp(-ax^2)dx = \frac{1}2 \sqrt{\frac{\pi}{a}}$$  I have to calculate   $$\int_{0}^\infty \exp \left(-y^2 - \frac{c^2}{y^2} \right)dy$$  So I think we need to use a u-substitution but i can't find something...  Is someone have an idea ? :)  Thanks for your answers  Mathieu
30169
I ran across an interesting integral and I am wondering if anyone knows where I may find its derivation or proof.  I looked through the site. If it is here and I overlooked it, I am sorry.  $$\displaystyle\frac{1}{\int_{-\infty}^{\infty}\frac{1}{(e^{x}-x)^{2}+{\pi}^{2}}dx}-1=W(1)=\Omega$$  $W(1)=\Omega$ is often referred to as the Omega Constant. Which is the solution to   $xe^{x}=1$.  Which is $x\approx .567$  Thanks much.   EDIT:  Sorry, I had the integral written incorrectly. Thanks for the catch.  I had also seen this:  $\displaystyle\int_{-\infty}^{\infty}\frac{dx}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+W(1)}=\frac{1}{1+\Omega}\approx .638$  EDIT:  I do not what is wrong, but I am trying to respond, but can not. All the buttons are unresponsive but this one. I have been trying to leave a greenie and add a comment, but neither will respond. I just wanted you to know this before you thought I was an ingrate. Thank you. That is an interesting site.
109962
I've been grappling for a few hours with this problem ($b$ is a parameter)  $\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}$  Tried direct integration by parts, differentiation w.r.t. $b$, then integrating... all to no avail. I think the second approach is the way... Pls help
43297
Can anyone give me a hint to evaluate this integral?  $$\int_0^\infty \frac{dx}{1+x^4}$$  I know it will involve the gamma function, but how?
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!
33935
Show that $$\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx=\frac{\pi^2}{4} \operatorname{sech}^2\left(\frac{a\pi}{2}\right) $$   I think we can solve it by contour integration, but how? and it would be better if there were two ways, the first with real analysis and the second with complex analysis.
107435
In , I used straightforward contour integration to evaluate the integral $$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^a}{1+x^2}\,dx=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)}$$for $|a|&lt;1$.   An alternative approach is to enforce the substitution $x\to e^x$ to obtain  $$\begin{align} \int_0^\infty \frac{x^a}{1+x^2}\,dx&amp;=\int_{-\infty}^\infty \frac{e^{(a+1)x}}{1+e^{2x}}\,dx\\\\ &amp;=\int_{-\infty}^0\frac{e^{(a+1)x}}{1+e^{2x}}\,dx+\int_{0}^\infty\frac{e^{(a-1)x}}{1+e^{-2x}}\,dx\\\\ &amp;=\sum_{n=0}^\infty (-1)^n\left(\int_{-\infty}^0 e^{(2n+1+a)x}\,dx+\int_{0}^\infty e^{-(2n+1-a)x}\,dx\right)\\\\ &amp;=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+a}+\frac{1}{2n+1-a}\right)\\\\ &amp;=2\sum_{n=0}^\infty (-1)^n\left(\frac{2n+1}{(2n+1)^2-a^2}\right) \tag 1\\\\ &amp;=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 2 \end{align}$$  Other possible ways forward include writing the integral of interest as  $$\begin{align} \int_0^\infty \frac{x^a}{1+x^2}\,dx&amp;=\int_{0}^1 \frac{x^{a}+x^{-a}}{1+x^2}\,dx \end{align}$$  and proceeding similarly, using $\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$.     Without appealing to complex analysis, what are other approaches one can use to evaluate this very standard integral?   EDIT:     Note that we can show that $(1)$ is the partial fraction representation of $(2)$ using Fourier series analysis.  I've included this development for completeness in the appendix of the solution I posted on .