new challenging integrals

**galactus** · January 11th 2009, 01:01 PM

Here's two integrals I ran into I thought y'all might like a go. Perhaps they are cliche?.

Show that:

This one has its own name. The Ahmed Integral.

$\int_{0}^{1}\frac{tan^{-1}}{x(x^{2}+1)}dx=\frac{\text{Catalan}}{2}+\frac{\ pi}{8}ln(2)$

In case, $\text{Catalan}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}\approx .915....$

The other:

$\int_{0}^{\infty}\frac{sin(x)}{x^{p}}dx=\frac{\sqr t{\pi}{\Gamma}(1-\frac{p}{2})}{2^{p}{\Gamma}(\frac{1}{2}+\frac{p}{2 })}$
or some equivalent form.

I thought these were cool

.

**Krizalid** · January 11th 2009, 03:05 PM

Originally Posted by galactus

Show that:

This one has its own name. The Ahmed Integral.

$\int_{0}^{1}\frac{tan^{-1}}{x(x^{2}+1)}dx=\frac{\text{Catalan}}{2}+\frac{\ pi}{8}ln(2)$

In case, $\text{Catalan}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}\approx .915....$

Your integral equals $\frac{1}{2}\left( \int_{0}^{1}{\frac{\ln \left( 1+{{y}^{2}} \right)}{{{y}^{2}}+1}\,dy}-\ln 2\int_{0}^{1}{\frac{dy}{{{y}^{2}}+1}} \right)-\int_{0}^{1}{\frac{\ln y}{{{y}^{2}}+1}\,dy}$

after some double integration manipulations. Those integrals are not hard to find, from there one gets the result.

**galactus** · January 11th 2009, 04:13 PM

Hey Kriz:

You are better at this than me. That's for sure. I have to ask. How in the world did you get that ln thing from that integral with arctan?.

I can do each of those integrals you have, but I never saw it in terms of ln.

I started out using the series for arctan, but the x^2+1 in there caused me a fit.

Using $tan^{-1}(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}$ .

$\int_{0}^{1}\frac{1}{x^{2}+1}\sum_{k=0}^{\infty}\f rac{(-1)^{k}x^{2k}}{2k+1}$

$\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}\frac{x^{2k}}{x^{2}+1}$ ..............[3]

I can see $\int x^{2k} dx=\frac{x^{2k+1}}{2k+1}$ , which

when multiplied by the existing 2k+1 will result in the Catalan, but I am

getting hung up because of the x^2+1.

Now, $\sum_{k=0}^{\infty}(-1)^{k}x^{2k}=\frac{1}{1+x^{2}}$ .

Then we can write $\int_{0}^{1}\sum_{k=0}^{\infty}(-1)^{k}x^{2k}\cdot x^{2k}$

$\int_{0}^{1}x^{2k}\sum_{k=0}^{\infty}(-1)^{k}x^{2k}$

When we integrate, we have $\frac{x^{2k+1}}{2k+1}\sum_{k=0}^{\infty}(-1)^{k}x^{2k}$

When multiplied with [3], we get the Catalan.

But what about the rest?. I know it;s there. I can smell it.

I am making a stupid mistake. I just know it. Every time I tried this thing I ended up in a dead end.

You probably see it right off, though.

**Krizalid** · January 11th 2009, 04:16 PM

$\int_{0}^{1}{\frac{\arctan x}{x\left( {{x}^{2}}+1 \right)}}\,dx=\int_{0}^{1}{\int_{0}^{x}{\frac{dy\, dx}{x\left( {{x}^{2}}+1 \right)\left( {{y}^{2}}+1 \right)}}}=\int_{0}^{1}{\int_{y}^{1}{\frac{dx\,dy} {x\left( {{x}^{2}}+1 \right)\left( {{y}^{2}}+1 \right)}}}.$

That leads the integrals I left before.

**galactus** · January 11th 2009, 04:20 PM

Oh, OK. Good egg. I see now what you done. Clever indeed.

I also know that $\int_{0}^{\frac{\pi}{4}}ln(1+tan(x))dx=\frac{\pi}{ 8}ln(2)$

Which is part of our solution. May have to look into how to transform it. I bet there is a connection.

Also,this integral becomes $\int_{0}^{\frac{\pi}{4}}xcot(x)dx$ , but the Pi/4 makes it more difficult than the usual Pi/2 for this integral.

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