I have been trying to solve this question with the simple methods I knew but couldn't come to the solution. I would appreciate a bit of hint to solve this integral problem?  $\;\displaystyle\int\frac {e^x}x\,dx$     Edit : Can't it be solved even if certain upper and lower limits are involved?
89875
I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong?  So, I'm trying to determine: $$ \int{\frac{e^x}{x}} \, dx $$  Integrate by parts, where $u = 1/x$, and $v \, ' = e^x$. Then $u \, ' = - 1/x^2$, and $v=e^x$. So,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \int{\frac{e^x}{x^2}} \, dx$$  Integrate by parts again, $u = 1/x^2$, $v \, ' = e^x$, so that $u \, ' = -2/x^3$ and $v=e^x$. So,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2\int{\frac{e^x}{x^3}} \, dx$$  Repeat this process ad infinitum to get,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2 \left( \frac{e^x}{x^3} + 3 \left( \frac{e^x}{x^4} + 4 \left( \frac{e^x}{x^5} + \, \cdots \right) \right) \right) $$  Expanding this gives,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots $$  And factoring that gives,  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left(  1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$  Now, considering the series itself, the ratio between the $n^{th}$ term and the $(n-1)^{th}$ term = $\Large \frac{n}{x}$. Eventually, $n$ will be larger than $x$, so the ratio between successive terms will be positive, so (assuming $x$ is positive), the series diverges, meaning (and I'm sure everybody will cringe upon seeing notation used like this), that:  $$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty $$
29157
Please give me some hints or solution for this limit. I had it on my exam and I'm curious how to solve it. $$ \lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x} $$
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?
63223
Given the integral $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$  How can I evaluate it? Thanks!
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.
13847
I am trying to compute this integral/limit, I don't feel like I have any good insight...   $$\lim_{n \to \infty} \int_0^{n^2} e^{-x^2}n\sin\frac{x}{n} \, dx.$$  I have tried to make a change of variable to get rid of the $n^2$, I changed to $X=\frac{x}{n^2}$ but got something even worse, I've tried to reach a situation where I could use a convergence theorem for Lebesgue Integrals,... I'm not sure I'm even on the right track!  Could you give me a hint on how to start this?  Thank you very much!
24079
Find the value of $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz$   The first idea I had was to apply Cauchy's integral formula. But can this be done? $e^{1/z^2}$ is not holomorphic on the interior of $|z|=2$. Still, if we would apply this we would obtain:  $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz=-2\pi ie$.  Another way I think, is to apply the residue theorem. For $1-z$ has a simple zero at $z=1$ and $e^{1/1^2}\neq 0$, then the residue of $e^{1/z^2}/(1-z)$ at $z=0$ is $-e$. Again we obtain $2\pi iRes(g;0)=-2\pi ie$, where $g=\frac{e^{1/z^2}}{1-z}$.  But again, it seems to me this is not right because $g$ has two singularities, at $z=1$ and $z=0$. What we might do about $z=0$? The answer is not $-2\pi i e$?
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.
52543
I have no idea how to start, it looks like integration by parts won't work.  $$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$  If someone could shed some light on this I'd be very thankful.
113629
I came across this integral $\int e^{x^n} dx $ and was just wondering how you go about solving it. In this  there approach is to let $y = -x^n$, then $$ dy = -nx^{n-1} dx = -n(-y)^{\frac{n-1}{n}} \ dx \ \Rightarrow \ dx = -\frac{1}{n}(-y)^{\frac{1-n}{n}} \ dy, $$ then sub into the integral, we have $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n}\int_{-\infty}^\infty  y^{\frac{1}{n}-1} e^{-y} \ dy, $$ then they referenced the primary definition of incomplete gamma function, $$ \Gamma(a,y) = \int_y^\infty t^{a-1} e^{-t} \ dt, $$ and claim that we can write our integral in terms of $\Gamma$ function, and the end result is $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, y). $$ Can anyone fill in the missing step for me? Our integral does look very similar to the gamma function, but what happened to the lower bound of the integral? In our integral the lower bound is $-\infty$; but to equate it to the gamma function we need the lower bound to be $y$.   Anyway substitute back the definition of $y$ we have the final result $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, -x^n). $$ I̶'̶m̶ ̶g̶u̶e̶s̶s̶i̶n̶g̶ ̶t̶h̶e̶i̶r̶ ̶a̶n̶s̶w̶e̶r̶ ̶i̶s̶ ̶c̶o̶r̶r̶e̶c̶t̶ ̶b̶e̶c̶a̶u̶s̶e̶ ̶M̶a̶t̶h̶e̶m̶a̶t̶i̶c̶a̶ ̶a̶l̶s̶o̶ ̶g̶i̶v̶e̶s̶ ̶t̶h̶e̶ ̶s̶a̶m̶e̶ ̶r̶e̶s̶u̶l̶t̶.̶ ̶B̶u̶t̶ ̶a̶n̶o̶t̶h̶e̶r̶ ̶t̶h̶i̶n̶g̶ ̶I̶ ̶d̶o̶n̶'̶t̶ ̶u̶n̶d̶e̶r̶s̶t̶a̶n̶d̶ ̶i̶s̶ ̶w̶h̶y̶ ̶d̶o̶e̶s̶ ̶t̶h̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶l̶ ̶d̶e̶p̶e̶n̶d̶s̶ ̶o̶n̶ ̶$̶x̶$̶?̶ ̶Y̶o̶u̶'̶d̶ ̶t̶h̶i̶n̶k̶ ̶$̶x̶$̶ ̶s̶h̶o̶u̶l̶d̶ ̶b̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶t̶e̶d̶ ̶o̶v̶e̶r̶.̶ Any help is appreciated, thanks.
110690
How do I start with evaluating this-     $$\int\frac{1+x^4}{(1-x^4)^{3/2}}dx$$   What should be my first attempt at this kind of a problem where-   The denominator and numerator are of the same degree Denominator involves fractional exponent like $3/2$.   Note:I am proficient with all kinds of basic methods of evaluating integrals.
75017
I am given the following problem which I have problem to know where to even start:     The question: $\lim_{x\to 0} \frac{\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt} {bx-esinx}=\frac{1}{\pi}$ The part where I will like to know where to start:  $\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt$   I appreciate suggestions on how I may get about solving the integration part before I move on to solve the limit question as a whole.
80733
Problem :   Let $n$ be a positive integer, for a real number $x$, let $[x]$ denote the largest integer not exceeding x and $\{x\} =x -[x]$ Then $$\int^{n+1}_1 \frac{(\{x\})^{[x]}}{[x]}dx$$ is equal to   (a) $\log_e(n)$   (b)  $\frac{1}{n+1}$   (c)  $\frac{n}{n+1}$   (d)  $1+\frac{1}{2}+\cdots +\frac{1}{n}$   Please suggest how to proceed in such cases not getting any clue how to proceed in cases of fractional part and greatest integral cases. please guide will be of great help thanks.
116269
$$\int_0^1\sqrt\frac x{1-x}\,dx$$ I saw in my book that the solution is $x=\cos^2u$ and $dx=-2\cos u\sin u\ du$. I would like to see different approaches, can you provide them?
74195
Good afternoon, I need to find:  $$\int_0^\infty\frac{x^n}{ax^2+bx+c}\space dx$$  I've no idea about any clue what so ever.
35482
Although I have known that $\displaystyle\int_0^\infty {{\sin x} \over x} \, dx = {\pi  \over 2}$, I have no idea how to work out $\displaystyle\int_0^{ + \infty } {{\cos x} \over x} \, dx$. How can I?
116992
During my studying to integration I find this integration.  So I tried to prove but I got stuk.  So I need help in this integration.  $$\displaystyle\int_{-\infty}^{\infty} \frac{x \sin (x)}{1+x^2}  dx = \frac {\pi}{e} $$