MHB How can I simplify this integral using integration by parts?

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The integral I(x) is defined as I(x) = ∫₀² (1+t) e^(x cos[π(t-1)/2]) dt, and the goal is to show that I(x) approximates to 4 + (8/π)x + O(x²) as x approaches 0. To simplify the integral, the McLaurin series expansion is utilized, leading to the evaluation of I(0) = 4 and the derivative I'(0) = ∫₀² cos[π(t-1)/2](1+t) dt, which simplifies to (8/π). The discussion emphasizes the use of differentiation under the integral sign to derive I'(x). Overall, the method clarifies how to handle the exponential term in the integral using integration by parts and series expansion techniques.
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Consider the integral
\begin{equation}
I(x)=\int^{2}_{0} (1+t) e^{xcos[\pi (t-1)/2]} dt
\end{equation}
show that
\begin{equation}
I(x)= 4+ \frac{8}{\pi}x +O(x^{2})
\end{equation}
as $x\rightarrow0$.=> Using integration by parts, but its too complicated for me because of huge exponential term.
please help me.
 
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First represent the integrand as $f(t)+xg(t)+O(x^2)$.
 
grandy said:
Consider the integral
\begin{equation}
I(x)=\int^{2}_{0} (1+t) e^{xcos[\pi (t-1)/2]} dt
\end{equation}
show that
\begin{equation}
I(x)= 4+ \frac{8}{\pi}x +O(x^{2})
\end{equation}
as $x\rightarrow0$.=> Using integration by parts, but its too complicated for me because of huge exponential term.
please help me.

If You expand I(x) in McLaurin series You have...

$\displaystyle I(x) = I(0) + I^{\ '} (0)\ x + \mathcal{O} (x^{2})\ (1)$

The first term is...

$\displaystyle I(0) = \int_{0}^{2} (1 + t)\ d t = 4\ (2)$

The $\displaystyle I^{\ '} (x)$ can be obtained deriving into the integral and is...

$\displaystyle I^{\ '} (0) = \int_{0}^{2} \cos [\frac{\pi}{2}\ (t-1)]\ (1 + t)\ d t = - \frac{2}{\pi^{2}}\ | \pi\ (1+ t)\ \cos ( \frac{\pi}{2}\ t )- 2\ \sin ( \frac{\pi}{2}\ t) |_{0}^{2} = \frac{8}{\pi}\ (3) $

Kind regards

$\chi$ $\sigma$
 
That was so clear. you made the solution so easy to understand. Thank you very much sir.
 
The $\displaystyle I^{\ '} (x)$ can be obtained deriving into the integral and is...

$\displaystyle I^{\ '} (0) = \int_{0}^{2} \cos [\frac{\pi}{2}\ (t-1)]\ (1 + t)\ d t $

how did you get the above term or $I'(x)$ term?
please clarify me. other than that everything is perfect.
 
grandy said:
The $\displaystyle I^{\ '} (x)$ can be obtained deriving into the integral and is...

$\displaystyle I^{\ '} (0) = \int_{0}^{2} \cos [\frac{\pi}{2}\ (t-1)]\ (1 + t)\ d t $

how did you get the above term or $I'(x)$ term?
please clarify me. other than that everything is perfect.

chisigma used this: Differentiation under the integral sign - Wikipedia, the free encyclopedia

He differentiated wrt $x$.
 
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