How to prove the following relation if $a&gt;0$ $$ \int_0^\infty \frac{\sin(ax)}{x^b}dx\,=\,\frac{1}{2}\pi\,a^{b-1}\frac{1}{\Gamma(b)\sin(\frac{b\pi}{2})}, \,\,\,\, 0&lt; \operatorname{Re} b &lt;2 $$
52200
How can I calculate the following integral:   $$\int_0^\infty x ^{-\frac{1}{3}}\sin x \, dx$$   WolframAlpha gives me $$ \frac{\pi}{\Gamma\Big(\frac{1}{3}\Big)}$$  How does WolframAlpha get this?  I don't understand how we can rearrange the formula in order to apply the gamma-function here. Any helpful and detailed hint/answer is appreciated.
103744
So, I was recently working on the Sine Fresnal integral and was curious whether we could generalise for any Real Number, i.e.   $$I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx$$  I have formed a solution that I'm uncomfortable with and was hoping for qualified eyes to have a look over.   So, the approach I took was to employ Complex Numbers (I forget the name(s) of the theorem that allows this).   But   $$\sin\left(x^n\right) = \Im\left[-e^{-ix^n}\right]$$  And so, n  $$ I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx = \Im\left[\int_{0}^{\infty} -e^{-ix^n}\:dx \right]= -\Im\left[\int_{0}^{\infty} e^{-\left(i^{\frac{1}{n}}x\right)^{n}}\:dx \right]$$  Applying a change of variable $u = i^{\frac{1}{n}}x$ we arrive at:  \begin{align} I &amp;= -\Im\left[i^{-\frac{1}{n}}\int_{0}^{\infty} e^{-u^{n}}\:du \right] \\   &amp;= -\Im\left[i^{-\frac{1}{n}}\frac{\Gamma\left(\frac{1}{n}\right)}{n} \right]\\   &amp;= \sin\left(\frac{\pi}{2n}\right)\frac{\Gamma\left(\frac{1}{n}\right)}{n} \end{align}  My area of concern is in the substitution. As $i^{-\frac{1}{n}} \in \mathbb{C}$, I believe the limits of the integral should have been from $0$ to $i^{-\frac{1}{n}}\infty$. Is that correct or not?  I'm also struggling with bounds on $n$ for convergence. Is this expression valid for all $n\in\mathbb{R}$  Any guidance would be greatly appreciated
90172
if $A_n \longrightarrow \infty $ and $B_n \longrightarrow \infty $  then $(A_n+B_n) \longrightarrow  \infty$.  How do you prove it?
87619
Show  $$\int_0^\infty \frac{\ln(x)}{e^x} = -\gamma$$ (gamma is Euler-Mascheroni constant).  Can anyone please prove this result?  Also  $$ \int_0^\infty \frac{\left( \ln(x) \right)^2}{e^x}\mathrm dx. $$
54941
$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$  Anyone an idea on how to prove this?
3970
I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$ What do you think?  It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this .
21748
I am having issues solving these problems: I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.     Show that for $a,b\in (-\infty,+\infty)$,   the following inequalities apply:   a) $2|ab| \leq a^2 + b^2$  b) $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$
121537
To prove: $$x-\frac{x^2}{2}&lt;\ln(1+x)&lt;x-\frac{x^2}{2(1+x)},\quad\forall x&gt;0$$  I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0.   Is there a cleaner / more elegant way of achieving the same?
90923
The  gives a simple formula for it in terms of the gamma function.  Using that and the fact that $\Gamma(n+1)=n!$, I can prove the following formula: $$ \begin{eqnarray*} \frac{a!b!}{(a+b+1)!} &amp; = &amp; \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+1+b+1)}\\  &amp; = &amp; B(a+1,b+1)\\  &amp; = &amp; \int_{0}^{1}t^{a}(1-t)^{b}dt\\  &amp; = &amp; \int_{0}^{1}t^{a}\sum_{i=0}^{b}\binom{b}{i}(-t)^{i}dt\\  &amp; = &amp; \int_{0}^{1}\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}t^{a+i}dt\\  &amp; = &amp; \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\int_{0}^{1}t^{a+i}dt\\  &amp; = &amp; \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\left[\frac{t^{a+i+1}}{a+i+1}\right]_{t=0}^{1}\\  &amp; = &amp; \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\\ b! &amp; = &amp; \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(a+b+1)!}{a!(a+i+1)} \end{eqnarray*} $$ This last formula involves only natural numbers and operations familiar in combinatorics, and it feels very much as if there should be a combinatoric proof, but I've been trying for a while and can't see it. I can prove it in the case $a=0$: $$ \begin{eqnarray*}  &amp;  &amp; \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(b+1)!}{0!(i+1)}\\  &amp; = &amp; \sum_{i=0}^{b}(-1)^{i}\frac{b!(b+1)!}{i!(b-i)!(i+1)}\\  &amp; = &amp; b!\sum_{i=0}^{b}(-1)^{i}\frac{(b+1)!}{(i+1)!(b-i)!}\\  &amp; = &amp; b!\sum_{i=0}^{b}(-1)^{i}\binom{b+\text{1}}{i+1}\\  &amp; = &amp; b!\left(1-\sum_{i=0}^{b+1}(-1)^{i}\binom{b+\text{1}}{i}\right)\\  &amp; = &amp; b! \end{eqnarray*} $$ Can anyone see how to prove it for arbitrary $a$? Thanks!
31979
Let $f(\sin x)$ be a given function of $\sin x$.   How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$?
7969
How to prove      $$I=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G$$   Where $ \operatorname{Li}_3(x)$ is the the trilogarithm function and $G$ is the Catalan constant.  Variant approaches are appreciated.
18509
I've been looking at  $$\int\limits_0^\infty  {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$  It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:  $$\displaystyle \int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$  $$\int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$  $$\int\limits_0^\infty  {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$  So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $\dfrac{1}{4}$, $\dfrac{1}{3}$ and $\dfrac{2}{5}$. Is there any way of finding it? I'll work on it and update with any ideas.    UPDATE:  The integral reduces to finding   $$\int\limits_{ - \infty }^\infty  {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt} $$  With $a =\dfrac{n+1}{m}$ which converges only if   $$0 &lt; a &lt; 1$$  Using series I find the solution is     $$\sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} $$
109169
This might be a silly question, but I'm reading  about differentiation under the integral sign, and I'm stumped by something that's written early on. The author is giving a derivation of the formula for $n!$ in terms of the gamma function. He shows how you can get  $$\frac{n!}{t^{n+1}} = \int_0^{\infty}x^ne^{-tx}\,dx$$  by differentating under the integral sign of $\int_0^{\infty}e^{-tx}dx$. He then says that the above "immediately implies" the formula  $$n! = \int_0^\infty x^ne^{-x}\,dx.$$  However, I can't for the life of me see how this follows. Multiplying the first equation by $t^{n+1}$ gives $n! = t^{n+1}\int_0^{\infty}x^ne^{-tx}$, so apparently  $$t^{n+1}\int_0^\infty x^ne^{-tx} \, dx = \int_0^\infty x^ne^{-x} \, dx,$$  but I don't see how this is true. Can anyone explain this?
114669
I am trying to solve:  $$\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^2 + a^2}dx$$  and show that it equals $\pi e^{-a}$ for all $a &gt; 0$.  I translate it to its complex equivalent:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz - \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$  where C is a half circle with radius R, above the real axis, and $\gamma$ is only the curved (top part) of the half circle.  I've shown that $$\int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$ tends to 0 as $R \rightarrow \infty$.  So now:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz$$  By residue theorem,   $$\int_{C}\frac{z\sin{z}}{z^2 + a^2}dz = 2\pi i \cdot res_{ai}f$$  as $ai$ is the pole of $f$ that lies above the real axis.  $$2\pi i \cdot res_{ai}f = 2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z sin{z}}{z^2 + a^2}$$  By L'Hospital's rule, this is equivalent to:  $$2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z^2\cos{z} + 2z\sin{z} - zai\cos{z} - ai\sin{z}}{2z} = 2\pi i \cdot \frac{\sin{ai}}{2} = \pi i \sin{ai}$$  But how is $\pi i \sin{ai}$ equivalent to $\pi e^{-a}$? Did I make a mistake?
23115
$(i)$ $a&gt;0,b&gt;0,p&gt;0,q&gt;0$   $(ii)$ $\frac{1}{p}+\frac{1}{q}=1$     How can we prove it?   ${a}{b}≤\frac{a^p}{p}+\frac{b^q}{q}$
26170
I am trying to compute the following integral using the Residue Theorem but am quite stuck: $$\int_0^\infty \frac{\sin^2x}{x^2}dx$$ I have tried applying Jordan's lemma, having written $\sin(x)$ as $\dfrac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}$, to not much avail. I have also tried using a rectangular integration path but it didn't get me far. I'd be grateful for an insightful advice.
87602
Let     $$ 1&lt; \alpha_1&lt;\alpha_2&lt;......&lt;\alpha_n&lt; \pi/2$$ then prove that $$ tan(\alpha_1)&lt; \frac{sin(\alpha_1)+sin(\alpha_2)+.....+sin(\alpha_n)}{cos(\alpha_1)+cos(\alpha_2)+......+cos(\alpha_n)} &lt; tan(\alpha_n)$$   I tried to solve the problem but I am not actually able to get a well established answer or you can say I am not even able to advance ahead to solve this one. It tried to use AM-GM inequality, but its didn't work as I couldn't establish any direct relation.