\( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
- Tolaso J Kos
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\( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
For the values of \(a \) for which the integral converges , evaluate : $$\int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx$$
HINT
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Re: \( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
Given \(\displaystyle \int_{0}^{1}\frac{1+x^a}{(1+x)^{a+2}}dx = \int_{0}^{1}\left(1+x\right)^{-a-2}dx+\int_{0}^{1}\frac{x^a}{(1+x)^{a+2}}dx\)
Now Let \(\displaystyle I = \int_{0}^{1}(1+x)^{-a-2}dx = -\frac{1}{(a+1)}\cdot \frac{1}{2^{a+1}}+\frac{1}{a+1}\)
Similarly Let \(\displaystyle J = \int_{0}^{1}\frac{x^a}{(1+x)^{a+2}}dx\), Let \(\displaystyle (1+x) = t\;,\) Then \(\displaystyle dx = dt\) and changing limit
Integral Convert into \(\displaystyle J = \int_{1}^{2}\frac{(t-1)^a}{t^{a+2}}dt = \int_{1}^{2}\left(1-\frac{1}{t}\right)^{a}\cdot \frac{1}{t^2} dt\)
Now let \(\displaystyle \left(1-\frac{1}{t}\right) = u\;,\) and \(\displaystyle \frac{1}{t^2}dt = du\) and changing Limit, We get
Integral \(\displaystyle J = \int_{0}^{\frac{1}{2}}u^a da = \frac{1}{a+1}\cdot \frac{1}{2^{a+1}}\)
So \(\displaystyle \int_{0}^{1}\frac{1+x^a}{(1+x)^{a+2}}dx = I+J = -\frac{1}{(a+1)}\cdot \frac{1}{2^{a+1}}+\frac{1}{a+1}+\frac{1}{(a+1)}\cdot \frac{1}{2^{a+1}} = \frac{1}{a+1}\)
P.S. Apostolos J. Kos, Would you like to explain it using Beta Theorem., Thanks
Now Let \(\displaystyle I = \int_{0}^{1}(1+x)^{-a-2}dx = -\frac{1}{(a+1)}\cdot \frac{1}{2^{a+1}}+\frac{1}{a+1}\)
Similarly Let \(\displaystyle J = \int_{0}^{1}\frac{x^a}{(1+x)^{a+2}}dx\), Let \(\displaystyle (1+x) = t\;,\) Then \(\displaystyle dx = dt\) and changing limit
Integral Convert into \(\displaystyle J = \int_{1}^{2}\frac{(t-1)^a}{t^{a+2}}dt = \int_{1}^{2}\left(1-\frac{1}{t}\right)^{a}\cdot \frac{1}{t^2} dt\)
Now let \(\displaystyle \left(1-\frac{1}{t}\right) = u\;,\) and \(\displaystyle \frac{1}{t^2}dt = du\) and changing Limit, We get
Integral \(\displaystyle J = \int_{0}^{\frac{1}{2}}u^a da = \frac{1}{a+1}\cdot \frac{1}{2^{a+1}}\)
So \(\displaystyle \int_{0}^{1}\frac{1+x^a}{(1+x)^{a+2}}dx = I+J = -\frac{1}{(a+1)}\cdot \frac{1}{2^{a+1}}+\frac{1}{a+1}+\frac{1}{(a+1)}\cdot \frac{1}{2^{a+1}} = \frac{1}{a+1}\)
P.S. Apostolos J. Kos, Would you like to explain it using Beta Theorem., Thanks
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Re: \( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
It is known that \( \displaystyle {\rm B}(x, y)=\int_{0}^{\infty}\frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\,dt \). From that we can obtain another great formula for the Beta Function which is \( \displaystyle {\rm B}(x, y)=\int_{0}^{1}\frac{t^{x-1}+t^{y-1}}{\left ( 1+t \right )^{x+y}}\,dt \).jacks wrote:P.S. Apostolos J. Kos, Would you like to explain it using Beta Theorem. Thanks
Our integral obeys to that formula so it can be-rewritten as : $$\int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx={\rm B}\left ( 1, a+1 \right )=\frac{\Gamma (1)\Gamma (a+1)}{\Gamma (a+2)}=\frac{\Gamma (a+1)}{(a+1)\Gamma (a+1)}=\frac{1}{a+1}$$ which is the result Jacks got.
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Re: \( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
I have just one question: Apostolos' solution implies that the given integral converges for \( \alpha > -1 \), since he makes use of Gamma (and Beta) function. Does jacks' solution imply that the given integral converges for \( \alpha < -1 \) as well?
- Tolaso J Kos
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Re: \( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
Nickos good evening,Tsakanikas Nickos wrote:I have just one question: Apostolos' solution implies that the given integral converges for \( \alpha > -1 \), since he makes use of Gamma (and Beta) function. Does jacks' solution imply that the given integral converges for \( \alpha < -1 \) as well?
From my solution is clear that integral converges for \(a>-1 \) as you said and that is correct. It is kwown that the \( \Gamma \) function \( \displaystyle \Gamma (x)=\int_{0}^{\infty}t^{x-1}e^{-t}\,dt \) converges when \( \mathfrak{Re}(x)>0 \). So, in my solution we demand both \(a+2>0 \) and \(a+1 >0 \). These have common solution when \(a>-1 \).
Thus the conclusion is that the integral converges when \(a>-1 \). Now note that the given integral diverges when \(a\leq -1 \). For example if \(a=-2 \) the given integral takes the form:
$$\int_{0}^{1}\frac{1+x^{-2}}{\left ( 1+x \right )^{0}}\,dx=\int_{0}^{1}\left ( 1+\frac{1}{x^2} \right )\,dx=+\infty$$
Similarly it can be shown that the integral diverges for \(a\leq -1 \) because it behaves like \( \dfrac{1}{x^a} \).
As for Jack's solution I don't think it implies convergence on \( \mathbb{R} \setminus \{-1\} \). Jacks has just evaluated the given integral taking for granted that it converges. He did not take into account divergence issues. He "chose" suitable values of \(a \) that made the integral converge.
I don't know if this suits your fancy.
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Re: \( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
Ok! Thanks a lot!Tolaso J Kos wrote:I don't know if this suits your fancy.
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