I am practicing calculus section of GRE Math Subject test, and I can't figure out a way to do the following integral: $$\int_0^\infty \frac{\cos (ax) - \cos (bx)}{x}dx.$$  I absolutely have no clue how to do this. Can someone show me the solution explicitly? Thank you.
12632
How can I prove the Theorem of Frullani? I did not even know all the hypothesis that $f$ must satisfy, but I think that this are  Let $\,f:\left[ {0,\infty } \right) \to \mathbb R$ be a a continuously differentiable function such that $$ \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 0, $$ and let $ a,b \in \left( {0,\infty } \right)$. Prove that $$ \int\limits_0^{\infty}  {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}} {x}}dx  = f\left( 0 \right)\left[ {\ln \frac{b} {a}} \right] $$  If you know a more general version please give it to me )= I can´t prove it.
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.
110711
$$\int_{-\infty}^{\infty}{{\rm d}x \over 1 + x^{2n}}$$  How to calculate this integral? I guess I need to use residue. But I looked at its solution. But it seems too complicated to me. Thus, I asked here. Thank you for help.
45585
I'm self studying complex analysis. I've encountered the following integral:  $$\int_{- \infty}^{+ \infty} \frac{e^{ax}}{1+e^x} dx \text{ with } a \in \mathbb{R},\  0 \lt a \lt 1. $$  I've done the substitution $e^x = y$. What kind of contour can I use in this case ?
123571
I want to show that $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$, but is there a faster method than finding the contour and doing all computations?  Otherwise my idea is to do the substitution $x=e^t$, integral than changes to $\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt$. Next step is to take the contour $-r,r,r+i\pi,-r+i\pi$ and integrate over it...
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)
34144
Please help me to find the integral of   $$\int \frac{x \sin{x} \cos{x}}{(a^2 \sin^2 x + b^2 \cos^2 x)^2} dx $$  There is a problem in having  x in the numerator. Please guide me . Thanks in advance.
12582
My problem is to integrate this expression:  $$\int_0^{2\pi}\log(1-2r\cos x +r^2)dx.$$ where $r$ is any constant in $[0,1]$.  I know the answer is zero. Can you explain you idea to me or just prove that? Maybe you will use the "Cauchy integral theorem ".
85114
I'm having trouble computing the integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$ I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.  Any suggestions are welcome. Thanks.
56394
I do not know how to solve this integral and then expand the result in powers of $t$  $$(1-t)^{-1/2}\int_{-\infty}^{\infty}\frac{\exp\left(\frac{2x^2t}{1+t}\right)}{1+x^2}dx=?,~~~~|t|&lt;1.$$  any idea?
40780
Evaluate: $$\int_{-\pi/4}^{\pi/4}(\cos(t)+\sqrt{1+t^2}\cos(t)^3\sin(t)^3)dt$$  I have tried substituting $t=\tan(\theta)$ and breaking up the $\cos(t)^3$ using $\cos(t)^2=1-\sin(t)^2$, but I can't figure it out.  This question is #21 from the Math subject GRE form GR1268.
102137
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$   My approach :   Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then  $$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$  But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
53738
Here is my question  $$\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$$  I have tried it by substituting $x$ = $\frac{1}{t}$. I got the answer $0$ but the correct answer is $\pi log(2)$. Any suggestion would be appreciated.
85817
How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.
76330
I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique.   When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely.   But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.
72851
Could someone help with the following integration:   $$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$  So far I have done the following, but I am stuck:  I denoted $ y=-\cos x $ then: $$\begin{align*}&amp;\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&amp;= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&amp;=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$  Then I am really stuck. Could someone help me?
106957
$$\int_0^\pi x \ln(\sin (x))dx  $$   I tried integrating this by parts but I end up getting integral that doesn't converge, which is this $$ \int_0^\pi \dfrac{x^2\cos (x)}{\sin(x)} \ dx$$ So can anyone help me on this one?