# Thread: Integration problems!

1. ## Integration problems!

Find the continuous function $f(x)$ and constant number $a$ which satisfies $\int_0^x {f(u)\,du} = e^x - ae^{2x} \int_0^1 {f(u) \cdot e^{ - u} \,du}$

Evaluate $\int_0^\infty {\frac{x}
{{e^x - 1}}\,dx}$

Evaluate $\int_0^1 {\frac{{x - 1}}
{{\ln x}}\,dx}$

Evaluate $\int_0^\pi {xe^x \sin x\,dx}$

Evaluate $\int_0^\infty {\frac{{e^{ - x} - 2e^{ - 3x} + e^{ - 5x} }}
{{x^2 }}\,dx}$

Evaluate $\int_0^1 {\int_0^1 {\frac{{x^3 y^3 }}
{{1 - xy}}\,dx} \,dy}$

Evaluate $\int_0^1 {\frac{{\arcsin x}}
{x}\,dx}$

Prove that $2\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = \int_{ - 1}^1 {\frac{1}
{{\sqrt {1 - x^2 } }}\,dx}$

Evaluate $\sum\limits_{k = 0}^\infty {\left( {\frac{1}
{{3k + 1}} - \frac{1}
{{3k + 2}}} \right)}$

Getting rid of them I'm done! Thanks!

2. Originally Posted by Mess
Find the continuous function $f(x)$ and constant number $a$ which satisfies $\int_0^x {f(u)\,du} = e^x - ae^{2x} \int_0^1 {f(u) \cdot e^{ - u} \,du}$
Differentiate this with respect to $x$, to get:

$
f(x)=e^x - 2a e^{2x} \int_0^1 f(u) e^{-u} du
$

Now put $K=\int_0^1 f(u) e^{-u} du$, then:

$
f(x)=e^x - 2a K e^{2x}
$

which you substitute back into the definition of $K$ to finish.

RonL

3. Slightly different from CaptainBlack's solution.

Originally Posted by Mess
Find the continuous function $f(x)$ and constant number $a$ which satisfies $\int_0^x {f(u)\,du} = e^x - ae^{2x} \int_0^1 {f(u) \cdot e^{ - u} \,du}$
Set $x=0,$

$0 = 1 - a\int_0^1 {f(u) \cdot e^{ - u} \,du} \,\therefore \,a\int_0^1 {f(u) \cdot e^{ - u} \,du} = 1.$

Hence $\int_0^x {f(u)\,du} = e^x - e^{2x} \implies f(x) = e^x - 2e^{2x} .$

Now this is routine, it remains to compute $\int_0^1 {\left( {e^u - 2e^{2u} } \right)e^{ - u} \,du} = 3 - 2e.$

The rest follows.

4. Originally Posted by Mess
Evaluate $\int_0^\infty {\frac{x}
{{e^x - 1}}\,dx}$
Effortlessly, start by proving that (for $k>0$) $\frac{1}
{{k^2 }} = \int_0^\infty {xe^{ - kx} \,dx} .$

And from the geometric series

$\frac{1}
{{e^x - 1}} = e^{ - x} \cdot \frac{1}
{{1 - e^{ - x} }} = e^{ - x} \sum\limits_{n = 0}^\infty {e^{ - nx} } = \sum\limits_{k = 1}^\infty {e^{ - kx} } .$

Interchange the integral by the sum and we'll get

$\int_0^\infty {\frac{x}
{{e^x - 1}}\,dx} = \int_0^\infty {x\sum\limits_{k = 1}^\infty {e^{ - kx} } \,dx} = \sum\limits_{k = 1}^\infty {\int_0^\infty {xe^{ - kx} \,dx} } = \sum\limits_{k = 1}^\infty {\frac{1}
{{k^2 }}} = \frac{{\pi ^2 }}
{6}.$

5. This kind of integral is well known. Actually, the only trick we need to apply, it's to construct a double integral. (Of course, I'd never create a double integral without switch the integration order.)

Originally Posted by Mess
Evaluate $\int_0^1 {\frac{{x - 1}}
{{\ln x}}\,dx}$
$\frac{{x - 1}}
{{\ln x}} = \int_0^1 {x^u \,du} .$
So,

$\int_0^1 {\frac{{x - 1}}
{{\ln x}}\,dx} = \int_0^1 {\int_0^1 {x^u \,du} \,dx} = \int_0^1 {\underbrace {\int_0^1 {x^u \,dx} }_{\dfrac{1}
{{u + 1}}}\,du} = \ln 2.$

6. Originally Posted by Mess
Evaluate $\int_0^1 {\int_0^1 {\frac{{x^3 y^3 }}
{{1 - xy}}\,dx} \,dy}$
$\frac{1}
{{1 - xy}} = \sum\limits_{k = 0}^\infty {(xy)^k } .$
So,

$\int_0^1 {\int_0^1 {\frac{{x^3 y^3 }}
{{1 - xy}}\,dx} \,dy} = \int_0^1 {\int_0^1 {\sum\limits_{k = 0}^\infty {x^{k + 3} y^{k + 3} } \,dx} \,dy} = \sum\limits_{k = 0}^\infty {\int_0^1 {\int_0^1 {x^{k + 3} y^{k + 3} \,dx} \,dy} } .$

And finally (evaluation of double integral is easy)

$\sum\limits_{k = 0}^\infty {\frac{1}
{{(k + 4)^2 }}} = \sum\limits_{n = 4}^\infty {\frac{1}
{{n^2 }}} = \frac{{\pi ^2 }}
{6} - 1 - \frac{1}
{4} - \frac{1}
{9} = \frac{{\pi ^2 }}
{6} - \frac{{49}}
{{36}}.$

7. There's something to be said about plain, old, dull, numerical methods.

$\int_{0}^{1}\frac{sin^{1}(x)}{x}dx$

By the midpoint method, we get $\frac{{\pi}ln(2)}{2}$

8. Originally Posted by Mess
Evaluate $\int_0^1 {\frac{{\arcsin x}}
{x}\,dx}$
Well, this one is not so hard to solve.

$\arcsin x = \int_0^x {\frac{1}
{{\sqrt {1 - u^2 } }}\,du} .$
So,

$\int_0^1 {\frac{{\arcsin x}}
{x}\,dx} = \int_0^1 {\int_0^x {\frac{1}
{{x\sqrt {1 - u^2 } }}\,du} \,dx} = \int_0^1 {\underbrace {\int_u^1 {\frac{1}
{x}\,dx} }_{ - \ln u}\frac{1}
{{\sqrt {1 - u^2 } }}\,du} ,$
hence

$\int_0^1 {\frac{{\arcsin x}}
{x}\,dx} = - \int_0^1 {\frac{{\ln u}}
{{\sqrt {1 - u^2 } }}\,du} .$

Now substitute $u=\sin\alpha,$ the integral becomes

$- \int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } = \frac{\pi }
{2}\ln 2,$

which confirms galactus' answer.

You can check the last result here. (See post #5.)

9. Hey Krizmeister.

I mostly go for the easy out, but I tried working the arcsine one out using your methods, only a slight variation. I hammered it into:

$\int_{0}^{1}\frac{1}{\sqrt{1-u^{2}}}du\cdot\int_{1}^{2}\frac{1}{x}dx=\frac{ln(2 )\pi}{2}$.

10. Originally Posted by Mess
Prove that $2\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = \int_{ - 1}^1 {\frac{1}
{{\sqrt {1 - x^2 } }}\,dx}$
$\pi=\pi$ is a boring way to claim equality, so let's apply some of double integration.

$2\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = 2\int_0^1 {2\sqrt {1 - x^2 } \,dx} .$

Insert $2\sqrt {1 - x^2 } = \int_{x^2 }^1 {\frac{1}
{{\sqrt {1 - u} }}\,du}$
and switch integration order

$\int_0^1 {\int_{x^2 }^1 {\frac{1}
{{\sqrt {1 - u} }}\,du} \,dx} = \int_0^1 {\underbrace {\int_0^{\sqrt u } {dx} }_{\sqrt u }\frac{1}
{{\sqrt {1 - u} }}\,du} .$

Now substitute $u=\varphi^2,$ then the integral becomes

$2\int_0^1 {\frac{{\varphi ^2 }}
{{\sqrt {1 - \varphi ^2 } }}\,d\varphi } = \int_{ - 1}^1 {\frac{1}
{{\sqrt {1 - \varphi ^2 } }}\,d\varphi } - \int_{ - 1}^1 {\sqrt {1 - \varphi ^2 } \,d\varphi } .$

From here $\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = \int_{ - 1}^1 {\frac{1}
{{\sqrt {1 - x^2 } }}\,dx} - \int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} ,$

rearrange and we're done.

11. Evaluate $\int_0^\pi {xe^x \sin x\,dx}$
Good, old-fashioned parts is one way to go on this one. Though, you'll have to apply it several times.

Let $u=xe^{x}, \;\ dv=sin(x)dx, \;\ du=(x+1)e^{x}, \;\ v=-cos(x)$

$-xe^{x}cos(x)+\int{e^{x}cos(x)}dx+\int{xe^{x}cos(x) }dx$

Do the parts thing again:

$\int{xe^{x}sin(x)}dx=-xe^{x}cos(x)+e^{x}cos(x)+\int{e^{x}sin(x)}dx+\int{ xe^{x}cos(x)}dx$

Continue in this fashion and you should have a $-\int{xe^{x}sin(x)}dx$ on the right side. Add it to both sides and divide by 2 and you should end up with:

$\frac{xe^{x}sin(x)}{2}+\frac{e^{x}cos(x)}{2}-\frac{xe^{x}cos(x)}{2}$

12. Originally Posted by Mess
Evaluate $\sum\limits_{k = 0}^\infty {\left( {\frac{1}
{{3k + 1}} - \frac{1}
{{3k + 2}}} \right)}$
$\frac{1}
{{3k + 1}} - \frac{1}
{{3k + 2}} = \int_0^1 {\left( {x^{3k} - x^{3k + 1} } \right)\,dx} = \int_0^1 {x^{3k} \left( {1 - x} \right)\,dx}$

Interchange sum & integral

$\sum\limits_{k = 0}^\infty {\left( {\frac{1}
{{3k + 1}} - \frac{1}
{{3k + 2}}} \right)} = \sum\limits_{k = 0}^\infty {\left( {\int_0^1 {x^{3k} \left( {1 - x} \right)\,dx} } \right)} = \int_0^1 {\sum\limits_{k = 0}^\infty {\Big[x^{3k} \left( {1 - x} \right)\Big]\,dx} } .$

Hence $\sum\limits_{k = 0}^\infty {\left( {\frac{1}
{{3k + 1}} - \frac{1}
{{3k + 2}}} \right)} = \int_0^1 {\frac{{1 - x}}
{{1 - x^3 }}\,dx} = \int_0^1 {\frac{1}
{{1 + x + x^2 }}\,dx} .$

The rest follows.

13. Originally Posted by Mess
Evaluate $\int_0^\infty {\frac{{e^{ - x} - 2e^{ - 3x} + e^{ - 5x} }}
{{x^2 }}\,dx}$
See post #4 (the parameter), it's all you need to solve the problem. (The interchange of integration order is justified by Tonelli's Theorem, since the function is positive.)

14. $- \int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } = \frac{\pi }
{2}\ln 2,$
I wanted to show my method of doing this integral. It's been bugging me. No, I am not vying for Best Integrator. I am sure Big K has that in the bag. I, like K, just enjoy doing these integrals. They are like little puzzles to figure out. I see why Plato says it is rather pointless, but why not just for fun and not practicality?.

$\int_{0}^{\frac{\pi}{2}}ln(sin({\theta}))d{\theta}$

First, I use the identity: $sin{\theta}=2sin(\frac{\theta}{2})cos(\frac{\theta }{2})$

$I=\int_{0}^{\frac{\pi}{2}}ln(sin{\theta})d{\theta} =\int_{0}^{\frac{\pi}{2}}ln(2sin\frac{\theta}{2}co s\frac{\theta}{2})d{\theta}$

$=\int_{0}^{\frac{\pi}{2}}ln2d{\theta}+\int_{0}^{\f rac{\pi}{2}}ln(sin\frac{\theta}{2})d{\theta}+\int_ {0}^{\frac{\pi}{2}}ln(cos\frac{\theta}{2})d{\theta }$

Now, make the sub $u=\frac{\theta}{2}$, and we get:

$I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s in(u))du+2\int_{0}^{\frac{\pi}{4}}ln(cos(u))du$

Now, use the identity $cos(u)=sin(\frac{\pi}{2}-u)$

and $z=\frac{\pi}{2}-u$.

We get:

$\int_{0}^{\frac{\pi}{4}}ln(cos(u))du = -\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}ln(sin(z))dz = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(sin(z))dz$

So, we get:

$I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s in(u))du+2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(s in(u))du$

$=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{2}}ln(si n({\theta}))d{\theta}=\frac{\pi}{2}ln(2)+2I$

Solve for I and get:

$\boxed{I = -\frac{\pi}{2}ln(2)}$

Of course, the negative at the beginning results in a positive solution.