Is it true that $  \displaystyle \int_{10}^\infty  \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx$ diverge since $  \displaystyle \int_{10}^\infty  \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx &gt;   \displaystyle \int_{10}^\infty  \cfrac{e^x}{x^{1 + \frac{1}{x}}e^x }dx=\displaystyle \int_{10}^\infty  \cfrac{1}{x^{1 + \frac{1}{x}} }dx$  and $\displaystyle \int_{10}^\infty  \cfrac{1}{x^{1 + \frac{1}{x}} }dx$ diverges by $p$-test?
122232
I want to calculate $  \displaystyle \int_{10}^\infty  \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx = \lim_{R \to \infty}\int_{10}^R  \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)} \, dx $. I think that it will diverge by bounding it below with $\frac{1}{x}$. So how do we calculate this integral?
66694
I'm currently tackling the following infinite series in order to determine if it converges or not:  $$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{n}$$  What I've done is rearranged this to make use of the comparison test, shown below:  $$= \frac{\sqrt{n+1}-\sqrt{n}}{n} * (\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}})$$  $$= \frac{(n+1)-(n)}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n^{3/2}+\sqrt{n+1}} &lt; \frac{1}{\sqrt{n+1}} &lt; \frac{1}{\sqrt{n}} $$  However, $\frac{1}{n}$ is smaller than $\frac{1}{\sqrt{n}}$, therefore:  $$  \frac{1}{n}   &lt; \frac{1}{\sqrt{n}}$$   This implies divergence by the comparison test due to the fact $\frac{1}{n}$ diverges, and if a smaller series diverges any larger series must as well.  Is this right? If I've made any mistakes in any part, please tell me!
44494
$$\sum_{n=1}^\infty \frac 1{n!}$$   I think we may compare it with $\frac{1}{n^n}$, and say that $\frac{1}{n!}&gt;\frac{1}{n^n}$. But I am not sure of the convergence or divergence of $\frac{1}{n^n}$ itself, so this turns out to be useless.  What other way is there?
56056
The value of $\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx$  equals 1.    The series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ equals $\pi^2/6 \approx 1.66493$.   Shouldn't the area under the curve of the function be larger than the sum of the terms of the sequence?
16758
The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges.  My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values.  Via L'Hospital, we have the following:  $$\lim_{x \to 0} x^c\ln(1+x)^q = \infty \space \text{when} \space  c &lt; 0$$ $$\lim_{x \to 0} x^c\ln(1+x)^q = 0 \space \text{when} \space  c &gt; 0$$ $$\lim_{x \to \infty} x^c\ln(1+x)^q = \infty \space \text{when} \space  c &gt; 0$$ $$\lim_{x \to \infty} x^c\ln(1+x)^q = 0 \space \text{when} \space  c &lt; 0$$  Thus, $$\int_1^{\infty} x^p\ln(1+x)^q dx = \int_1^{\infty} x^{p-c}x^c\ln(1+x)^q dx &gt; \int_1^{\infty} x^{p-c} dx$$ which diverges when $p &gt; -1$.  Here, we set $c = \frac{1}{2}(p+1) &gt; 0$.  On the other hand, $$\int_0^1 x^p\ln(1+x)^q dx = \int_0^1 x^{p-c}x^c\ln(1+x)^q dx &gt; \int_0^1 x^{p-c} dx$$ which diverges when $p &lt; -1$.  Here, we set $c = \frac{1}{2}(p+1) &lt; 0$.  Thus, the integral diverges whenever $p \ne –1$.  Then, let $p = -1$.  $$\int_0^{\infty} \frac{1}{x}\ln(1+x)^q dx = \int_0^1 \frac{1}{x}\ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x}\ln(1+x)^q dx $$ $$ \ge \int_0^1 \frac{1}{x+1}\ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x+1}\ln(1+x)^q dx $$  The first summand diverges when $q &lt; -1$ and the second diverges when $q &gt; -1$.  Finally, let $ p = -1$ and $q = -1$.  The integral $\int_0^{\infty} \frac{1}{x\ln(1+x)} dx ≥ \int_0^{\infty} \frac{1}{(x+1)\ln(1+x)} dx$ diverges on both ends.  Did I make a mistake?
80207
$ \int^\infty_0 (x^n)(e^{-x})  dx  $ and show that it is equal to $(n!)$ ?  I know that if you differentiate $x^n$  infinitely you get $n!$ but I don't know how to prove?
69259
If $\displaystyle S=\sum^{n}_{k=0}(-1)^k\frac{1}{k+m+1}\binom{n}{k}$ and f $\displaystyle T=\sum^{m}_{k=0}(-1)^k\frac{1}{k+n+1}\binom{m}{k}$.Then $S-T$ is   Try:$$S=\sum^{n}_{k=0}(-1)^k\int^{1}_{0}x^{k+n}\binom{n}{k}dx=\int^{1}_{0}x^n\sum^{n}_{k=0}(-x)^k\binom{n}{k}dx$$  Could some help me, thanks
35840
$\displaystyle\int {\frac{dx}{( 1+\sqrt {x})\sqrt{(x-{x}^2)}}}$   $=\displaystyle\int\frac{(1-\sqrt x)}{(1+x)\sqrt{x-x^2}}\,dx$      $=\displaystyle\int\frac{(1-\sqrt x+x-x)}{(1+x)\sqrt{x-x^2}}\,dx$   $=\displaystyle\int\frac{\,dx}{\sqrt{x-x^2}}-\displaystyle\int\frac{\sqrt x(1+\sqrt x)}{\sqrt x(1+x)\sqrt{1-x}}\,dx$   I have tried the above integration a lot . But not able to solve further .   Please give me some hint how to proceed or some other method
87619
Show  $$\int_0^\infty \frac{\ln(x)}{e^x} = -\gamma$$ (gamma is Euler-Mascheroni constant).  Can anyone please prove this result?  Also  $$ \int_0^\infty \frac{\left( \ln(x) \right)^2}{e^x}\mathrm dx. $$
20193
Find $\displaystyle \int_{0}^{\infty} \dfrac{\ln(x)}{1+x^2}dx$.   How should I change the limits of integration to evaluate this?
39848
Does the following series converge or diverge:  $$\sum_{n=1}^\infty \frac{3^{1/\sqrt{n}}-1}{n}$$  Many thanks!
86270
We know that $\sum \frac{1}{n^p}$ is convergent for $p&gt;1$. However the series $\sum {\frac{1}{n^{1+1/n}}}$ is apparently divergent since $1+1/n$ tends to 1 as $n$ tends to infinity. But how to prove this? The root test fails expectedly and I haven't been able to find a smaller divergent series for comparison test. Can someone help? Thanks!
16471
I am struggling to show that $$\frac{1}{\sqrt{2\pi}} \cdot \left(\int_{- \infty}^{\infty} e^{itx} \cdot e^{-\frac{1}{2}x^2 } dx \right) = e^{-\frac{1}{2}t^2}$$  In the same paper we defined  $$F(t):=\left(\int_0^t e^{-x^2}\right)^2 \quad \text{ and} \quad  G(t):=\left( \int _0^1\frac{e^{-t^2(1+x^2)}}{1+x^2} \right)$$  I showed that both functions are differentiable with  $$F'= 2e^{-t^2} \cdot \int_0^t e^{-x^2}$$ $$G' = -2te^{-t^2} \cdot \int_0^1 e^{-x^2t^2}$$  It follows that $F' + G' = 0$ and $F + G = \frac{\pi}{4}$  With that I showed that $\int_{-\infty}^{\infty} e^{-x^2} = \sqrt{\pi}$.  Any help is appreciated.
117530
Use a comparison test to determine whether or not the following improper integrals converge or diverge $$\int_{2}^\infty \frac{1}{\ln(x)}dx.$$ I'm stuck thinking of a function to compare it to.
71393
Find a general formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^a e^{-bx}dx.$   I've been pondering about this question for some time. Obviously integration by parts is way too inefficient. I know there's recursion involved and maybe there's a way to find a general formula through intuition.
55255
I am particularly curious about exchanging the sum and integral in the third equation, if this is unjustified then please let me know.  We have that   $$\lim_{n \to \infty} n \int_{1}^{\infty}\frac{dx}{1+x^n} =\lim_{a \to \infty}\lim_{n \to \infty} n \int_1^{a}\frac{dx}{1+x^n} $$  Also   $$\frac{1}{1+x^n} = \frac{1}{x^n}\frac{1}{1+\frac{1}{x^n}} = \sum_{i=0}^{\infty}\bigg(\frac{-1}{x^n}\bigg)^i\frac{1}{x^n} =\sum_{i=0}^{\infty}(-1)^i \frac{1}{x^{(i+1)n}}$$  Since we are integrating for $x \geq 1$, the above series converges for every $x$ so we may integrate term-by-term to obtain   $$n \int_1^{a}\frac{1}{1+x^n}dx = n\sum_{i=0}^{\infty}\int_1^{a}\frac{1}{x^{(i+1)n}} \, dx$$  $$= n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-n\left[\frac{a^{1-n}}{n-1}-\frac{a^{1-2n}}{2n-1}+\cdots\right] $$  where the second term goes to zero as $n$ goes to infinity. Further, we have that   $$\lim_{n \to \infty}\frac{n}{nk-1} = \frac{1}{k}$$  so the first term becomes $1 -\frac{1}{2}+\frac{1}{3}-... = \ln(2)$. In conclusion we get  $$\lim_{a \to \infty}\lim_{n \to \infty} n \int_1^{a}\frac{dx}{1+x^n} =\lim_{a \to\infty} \ln(2) = \ln(2) $$  independently of $a$.
34849
I'm trying to show that the series $\sum_{n=1}^\infty\frac{n}{n^2+1}$ is divergent.   I'm trying to use a comparison test to do so. To do so, I first can see that $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{n=1}^\infty\frac{n}{n^2+n} = \sum_{n=1}^\infty\frac{1}{n+1}$ And I can use a change of variable $k = n+1$, so show $\sum_{n=1}^\infty\frac{1}{n+1} = \sum_{k=2}^\infty\frac{1}{k} = \infty$. Then since $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{k=2}^\infty\frac{1}{k}$ it must be that $\sum_{k=2}^\infty\frac{1}{k}$ also diverges to infinity. Is this a valid way to show it?