Solving and Plotting an integral function

Question

I am new in the use of Mathematica, so please forgive me..

I want to integrate and plot the following integral function:

$F(x)=\displaystyle\int_x^\infty\dfrac{t\sqrt{t^2-x^2}}{e^t-1}\text{d}t$:

where $x>0$ is the argument of my integral function.

Mathematica doesn't manage to do the calculation, not even if I substitute the x inside the square root with a number, e.g. 1.

Can anyone please help me? Thanks

Possible duplicate of What are the most common pitfalls awaiting new users? — MarcoB, Commented Apr 15, 2019 at 18:59

Ulrich Neumann · Accepted Answer · 2019-04-15 14:11:01Z

4

Your problem can be solved numerically:

F[x_?NumericQ] :=NIntegrate[(t Sqrt[t^2 - x^2])/(Exp[t] - 1), {t, x, Infinity}]
Plot[F[x], {x, 0, 10}]

answered Apr 15, 2019 at 14:11

Ulrich Neumann

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$\begingroup$ Thank you very much, it was silly I guess... Just a brief question: what is the meaning of using "?NumericQ" in the argument of the function? $\endgroup$
– Lele
Commented Apr 15, 2019 at 14:21
$\begingroup$ @EmanueleCopello - Look at the documentation for PatternTest. Any function which uses numeric techniques such as NIntegrate can only evaluate if its argument(s) is/are numeric, so the argument(s) should be restricted to being numeric. $\endgroup$
– Bob Hanlon
Commented Apr 15, 2019 at 16:16
$\begingroup$ Ok thank you. What if I have to evaluate it in this way: $F(x)=\sum_{j=1}^n\displaystyle\int_{x_j}^\infty\dfrac{t\sqrt{t^2-x_j^2}}{e^t-1}\text{d}t$ where $x_j$ should be a list of values I have to insert somehow... What is the best computational way to do it? $\endgroup$
– Lele
Commented Apr 16, 2019 at 10:22
$\begingroup$ The sum(!) doesn't depend on x and therefor doesn't define F[x]! $\endgroup$
– Ulrich Neumann
Commented Apr 16, 2019 at 10:31
$\begingroup$ Ok sorry, I have to specify that better.. My function is in fact this one: $F(T)=\displaystyle\sum_{j=1}^n\displaystyle\int_{m_j/T}^\infty\dfrac{t\sqrt{t^2-(m_j/T)^2}}{e^t-1}\text{d}t$ $\endgroup$
– Lele
Commented Apr 16, 2019 at 10:45

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